RD Sharma XI 2020 Volume 1 Solutions for Class 11 Science Math Chapter 19 - Arithmetic Progressions

Share with your friends

RD Sharma XI 2020 Volume 1 Solutions for Class 11 Science Math Chapter 19 Arithmetic Progressions are provided here with simple step-by-step explanations. These solutions for Arithmetic Progressions are extremely popular among class 11 Science students for Math Arithmetic Progressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 Volume 1 Book of class 11 Science Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 Volume 1 Solutions. All RD Sharma XI 2020 Volume 1 Solutions for class 11 Science Math are prepared by experts and are 100% accurate.

Page No 19.11:

Question 1:

Find:
(i) 10th term of the A.P. 1, 4, 7, 10, ...
(ii) 18th term of the A.P. $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2},$ ...
(iii) nth term of the A.P. 13, 8, 3, −2, ...

Answer:

(i) 1, 4, 7, 10...

We have:
a = 1
d = $4 - 1 = 3$

$a_{10} = a + (10 - 1) d [a_{n} = a + (n - 1) d] = a + 9 d = 1 + 9 \times 3 = 28$

(ii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}$ ...
We have:
$a = \sqrt{2} d = 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2}$

$a_{18} = a + (18 - 1) d [a_{n} = a + (n - 1) d] = a + 17 d = \sqrt{2} + 17 (2 \sqrt{2}) = \sqrt{2} + 34 \sqrt{2} = 35 \sqrt{2}$

(iii) 13, 8, 3, −2...
We have:
$a = 13 d = 8 - 13 = - 5$

$a_{n} = a + (n - 1) d = 13 + (n - 1) (- 5) = 13 - 5 n + 5 = 18 - 5 n$

Page No 19.12:

Question 2:

If the sequence < a_n > is an A.P., show that
a_m_+n +a_m_{− n} = 2a_m.

Answer:

Let the sequence < a_n > be an A.P. with the first term being A and the common difference being D.
To prove: a_m_+n +a_m_{− n} = 2a_m

LHS: a_m_+n +a_m_{− n

$= A + (m + n - 1) D + A + (m - n - 1) D {∵ a_{n} = a + (n - 1) d} = A + m D + n D - D + A + m D - n D - D = 2 A + 2 m D - 2 D . . . (i)$}

RHS: 2a_m

_{$= 2 [A + (m - 1) D] = 2 A + 2 m D - 2 D . . . (ii)$}

From (i) and (ii), we get:
LHS = RHS
Hence, proved.

Page No 19.12:

Question 3:

(i) Which term of the A.P. 3, 8, 13, ... is 248?
(ii) Which term of the A.P. 84, 80, 76, ... is 0?
(iii) Which term of the A.P. 4, 9, 14, ... is 254?

Answer:

(i) 3, 8, 13...
Here, we have:
a = 3
$d = (8 - 3) = 5$
$Let a_{n} = 248 \Rightarrow a + (n - 1) d = 248 \Rightarrow 3 + (n - 1) 5 = 248 \Rightarrow (n - 1) 5 = 245 \Rightarrow n - 1 = 49 \Rightarrow n = 50$

Hence, 248 is the 50th term of the given A.P.

(ii) 84, 80, 76...
Here, we have:
a = 84
$d = (80 - 84) = - 4$
$L e t a_{n} = 0 \Rightarrow a + (n - 1) d = 0 \Rightarrow 84 + (n - 1) (- 4) = 0 \Rightarrow (n - 1) (- 4) = - 84 \Rightarrow (n - 1) = 21 \Rightarrow n = 22$

Hence, 0 is the 22nd term of the given A.P.

(iii) 4, 9, 14...
Here, we have:
a = 4
$d = (9 - 4) = 5$
$L e t a_{n} = 254 \Rightarrow a + (n - 1) d = 254 \Rightarrow 4 + (n - 1) 5 = 254 \Rightarrow (n - 1) 5 = 250 \Rightarrow (n - 1) = 50 \Rightarrow n = 51$

Hence, 254 is the 51st term of the given A.P.

Page No 19.12:

Question 4:

(i) Is 68 a term of the A.P. 7, 10, 13, ...?
(ii) Is 302 a term of the A.P. 3, 8, 13, ...?

Answer:

(i) 7, 10, 13...
Here, we have:
a = 7
$d = (10 - 7) = 3 Let a_{n} = 68 \Rightarrow a + (n - 1) d = 68 \Rightarrow 7 + (n - 1) (3) = 68 \Rightarrow (n - 1) (3) = 61 \Rightarrow (n - 1) = \frac{61}{3} \Rightarrow n = \frac{61}{3} + 1 = \frac{64}{3}$

Since n is not a natural number.So, 68 is not a term of the given A.P.

(ii) 3, 8, 13...
Here, we have:
a = 3
$d = (8 - 3) = 5 Let a_{n} = 302 \Rightarrow a + (n - 1) d = 302 \Rightarrow 3 + (n - 1) 5 = 302 \Rightarrow (n - 1) 5 = 299 \Rightarrow (n - 1) = \frac{299}{5} \Rightarrow n = \frac{299}{5} + 1 = \frac{304}{5}$
Since n is not a natural number.So, 302 is not a term of the given A.P.

Page No 19.12:

Question 5:

(i) Which term of the sequence 24, $23 \frac{1}{4,} 22 \frac{1}{2,} 21 \frac{3}{4}$ ... is the first negative term?
(ii) Which term of the sequence 12 + 8i, 11 + 6i, 10 + 4i, ... is (a) purely real (b) purely imaginary?

Answer:

(i) 24, $23 \frac{1}{4,} 22 \frac{1}{2,} 21 \frac{3}{4}$ ...
This is an A.P.
Here, we have:
a = 24
$d = (23 \frac{1}{4} - 24) = (- \frac{3}{4}) Let the first negative term be a_{n} . Then, w e h a v e : a_{n} < 0 \Rightarrow a + (n - 1) d < 0 \Rightarrow 24 + (n - 1) (- \frac{3}{4}) < 0 \Rightarrow 24 - \frac{3 n}{4} + \frac{3}{4} < 0 \Rightarrow 24 + \frac{3}{4} < \frac{3 n}{4} \Rightarrow \frac{99}{4} < \frac{3 n}{4} \Rightarrow 99 < 3 n \Rightarrow n > 33$

Thus, the 34th term is the first negative term of the given A.P.

(ii)
(a) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
$d = (11 + 6 i - 12 - 8 i) = (- 1 - 2 i) Let the real term be a_{n} = a + (n - 1) d . a_{n} = (12 + 8 i) + (n - 1) (- 1 - 2 i) = (12 + 8 i) + (- n + 1 - 2 i n + 2 i) = 12 + 8 i - n + 1 - 2 i n + 2 i = (13 - n) + (8 - 2 n + 2) i = (13 - n) + (10 - 2 n) i a_{n} has to be real . ∴ (10 - 2 n) = 0 \Rightarrow n = 5$

(b) 12 + 8i, 11 + 6i, 10 + 4i...
This is an A.P.
Here, we have:
a = 12 + 8i
$d = (11 + 6 i - 12 - 8 i) = (- 1 - 2 i) Let the imaginary term be a_{n} = a + (n - 1) d a_{n} = (12 + 8 i) + (n - 1) (- 1 - 2 i) = (12 + 8 i) + (- n + 1 - 2 i n + 2 i) = 12 + 8 i - n + 1 - 2 i n + 2 i = (13 - n) + (8 - 2 n + 2) i = (13 - n) + (10 - 2 n) i a_{n} has to be imaginary . ∴ (13 - n) = 0 \Rightarrow n = 13$

Page No 19.12:

Question 6:

(i) How many terms are there in the A.P.
7, 10, 13, ... 43?
(ii) How many terms are there in the A.P.
$- 1, - \frac{5}{6}, - \frac{2}{3}, - \frac{1}{2}, . . ., \frac{10}{3} ?$

Answer:

(i) 7, 10, 13...43
Here, we have:
a = 7
$d = (10 - 7) = 3 a_{n} = 43$
Let there be n terms in the given A.P.

$Also, a_{n} = a + (n - 1) d \Rightarrow 43 = 7 + (n - 1) 3 \Rightarrow 36 = (n - 1) 3 \Rightarrow 12 = (n - 1) \Rightarrow 13 = n$

Thus, there are 13 terms in the given A.P.

(ii) $- 1, - \frac{5}{6}, - \frac{2}{3}, - \frac{1}{2}, . . ., \frac{10}{3}$

Here, we have:
a = $- 1$
$d = (\frac{- 5}{6} - (- 1)) = (1 - \frac{5}{6}) = \frac{1}{6} a_{n} = \frac{10}{3}$
Let there be n terms in the given A.P.

$Also, a_{n} = a + (n - 1) d \Rightarrow \frac{10}{3} = - 1 + (n - 1) \frac{1}{6} \Rightarrow \frac{13}{3} = (n - 1) \frac{1}{6} \Rightarrow 26 = (n - 1) \Rightarrow 27 = n$

Thus, there are 27 terms in the given A.P.

Page No 19.12:

Question 7:

The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.

Answer:

Here, a = 5, d = 3, a_n = 80
Let the number of terms be n.
Then, we have:

$a_{n} = a + (n - 1) d \Rightarrow 80 = 5 + (n - 1) 3 \Rightarrow 75 = (n - 1) 3 \Rightarrow 25 = (n - 1) \Rightarrow 26 = n$

Thus, there are 26 terms in the given A.P.

Page No 19.12:

Question 8:

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer:

Given:

$a_{6} = 19 \Rightarrow a + (6 - 1) d = 19 [a_{n} = a + (n - 1) d] \Rightarrow a + 5 d = 19 . . (1) And, a_{17} = 41 \Rightarrow a + (17 - 1) d = 41 [a_{n} = a + (n - 1) d] \Rightarrow a + 16 d = 41 . . (2) Solving the two equations, we get, 16 d - 5 d = 41 - 19 \Rightarrow 11 d = 22 \Rightarrow d = 2 Putting d = 2 in the eqn (1), we get : a + 5 \times 2 = 19 \Rightarrow a = 19 - 10 \Rightarrow a = 9$

We know:
$a_{40} = a + (40 - 1) d [a_{n} = a + (n - 1) d] = a + 39 d = 9 + 39 \times 2 = 9 + 78 = 87$

Page No 19.12:

Question 9:

If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

Answer:

Given:
$a_{9} = 0 \Rightarrow a + (9 - 1) d = 0 [a_{n} = a + (n - 1) d] \Rightarrow a + 8 d = 0 \Rightarrow a = - 8 d . . . (i)$

To prove:
$a_{29} = 2 a_{19}$

Proof:
$LHS : a_{29} = a + (29 - 1) d = a + 28 d = - 8 d + 28 d (From (i)) = 20 d$

$RHS : 2 a_{19} = 2 [a + (19 - 1) d] = 2 (a + 18 d) = 2 a + 36 d = 2 (- 8 d) + 36 d (From (i)) = - 16 d + 36 d = 20 d$

LHS = RHS
Hence, proved.

Page No 19.12:

Question 10:

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Answer:

Given:
$10 a_{10} = 15 a_{15} \Rightarrow 10 [a + (10 - 1) d] = 15 [a + (15 - 1) d] \Rightarrow 10 (a + 9 d) = 15 (a + 14 d) \Rightarrow 10 a + 90 d = 15 a + 210 d \Rightarrow 0 = 5 a + 120 d \Rightarrow 0 = a + 24 d \Rightarrow a = - 24 d . . . (i)$

To show:
$a_{25} = 0 \Rightarrow LHS : a_{25} = a + (25 - 1) d = a + 24 d = - 24 d + 24 d (From (i)) = 0 = RHS Hence, proved .$

Page No 19.12:

Question 11:

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer:

Given:

$a_{10 =} 41 \Rightarrow a + (10 - 1) d = 41 [a_{n} = a + (n - 1) d] \Rightarrow a + 9 d = 41 And, a_{18} = 73 \Rightarrow a + (18 - 1) d = 73 [a_{n} = a + (n - 1) d] \Rightarrow a + 17 d = 73 Solving the two equations, we get : \Rightarrow 17 d - 9 d = 73 - 41 \Rightarrow 8 d = 32 \Rightarrow d = 4 . . . (i) Putting the value in first equation, we get : a + 9 \times 4 = 41 \Rightarrow a + 36 = 41 \Rightarrow a = 5 . . . (i i)$

$a_{26} = a + (26 - 1) d [a_{n} = a + (n - 1) d] \Rightarrow a_{26} = a + 25 d \Rightarrow a_{26} = 5 + 25 \times 4 (From (i) and (ii)) \Rightarrow a_{26} = 5 + 100 = 105$

Page No 19.12:

Question 12:

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer:

Given:
$a_{24} = 2 a_{10} \Rightarrow a + (24 - 1) d = 2 [a + (10 - 1) d] \Rightarrow a + 23 d = 2 (a + 9 d) \Rightarrow a + 23 d = 2 a + 18 d \Rightarrow 5 d = a . . . (i)$

$To prove : a_{72} = 2 a_{34} LHS : a_{72} = a + (72 - 1) d \Rightarrow a_{72} = a + 71 d \Rightarrow a_{72} = 5 d + 71 d (From (i)) \Rightarrow a_{72} = 76 d$

$RHS : 2 a_{34} = 2 [a + (34 - 1) d] \Rightarrow 2 a_{34} = 2 (a + 33 d) \Rightarrow 2 a_{34} = 2 (5 d + 33 d) (Form (i)) \Rightarrow 2 a_{34} = 2 (38 d) \Rightarrow 2 a_{34} = 76 d$
∴ RHS = LHS
Hence, proved.

Page No 19.12:

Question 13:

If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.

Answer:

Given:
$a_{m + 1} = 2 a_{n + 1} \Rightarrow a + (m + 1 - 1) d = 2 [a + (n + 1 - 1) d] \Rightarrow a + m d = 2 (a + n d) \Rightarrow a + m d = 2 a + 2 n d \Rightarrow 0 = a + 2 n d - m d \Rightarrow n d = \frac{m d - a}{2} . . . (i)$

To prove:

$a_{3 m + 1} = 2 a_{m + n + 1} LHS : a_{3 m + 1} = a + (3 m + 1 - 1) d \Rightarrow a_{3 m + 1} = a + 3 m d RHS : 2 a_{m + n + 1} = 2 [a + (m + n + 1 - 1) d] \Rightarrow 2 a_{m + n + 1} = 2 (a + m d + n d) \Rightarrow 2 a_{m + n + 1} = 2 [a + m d + (\frac{m d - a}{2})] (From (i)) \Rightarrow 2 a_{m + n + 1} = 2 [\frac{2 a + 2 m d + m d - a}{2}]$

$\Rightarrow 2 a_{m + n + 1} = 2 [\frac{a + 3 m d}{2}] \Rightarrow 2 a_{m + n + 1} = a + 3 m d$

∴ LHS = RHS

Page No 19.12:

Question 14:

If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Answer:

Given:
nth term of the A.P. 9, 7, 5... is the same as the nth term of the A.P. 15, 12, 9...

$Considering 9, 7, 5 ... a = 9, d = (7 - 9) = - 2 n^{th} term = 9 + (n - 1) (- 2) [a_{n} = a + (n - 1) d] = 9 - 2 n + 2 = 11 - 2 n . . . (i)$

$Considering 15, 12, 9, ... a = 15, d = (12 - 15) = - 3 n^{th} term = 15 + (n - 1) (- 3) [a_{n} = a + (n - 1) d] = 15 - 3 n + 3 = 18 - 3 n . . . (ii)$

Equating (i) and (ii), we get:
$11 - 2 n = 18 - 3 n \Rightarrow n = 7$

Thus, 7th terms of both the A.P.s are the same.

Page No 19.12:

Question 15:

Find the 12th term from the end of the following arithmetic progressions:
(i) 3, 5, 7, 9, ... 201
(ii) 3, 8, 13, ..., 253
(iii) 1, 4, 7, 10, ..., 88

Answer:

(i) 3, 5, 7, 9...201
Consider the given progression with 201 as the first term and −2 as the common difference.
12th term from the end = $201 + (12 - 1) (- 2) = 179$

(ii) 3, 8, 13...253
Consider the given progression with 253 as the first term and −5 as the common difference.
12th term from the end = $253 + (12 - 1) (- 5) = 198$

(iii) 1, 4, 7, 10...88
Consider the given progression with 88 as the first term and −3 as the common difference.
12th term from the end = $88 + (12 - 1) (- 3) = 55$

Page No 19.12:

Question 16:

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

Answer:

Given:
Let the first term of the A.P. be a and the common difference be d.
$a_{4} = 3 a \Rightarrow a + (4 - 1) d = 3 a \Rightarrow a + 3 d = 3 a \Rightarrow 3 d = 2 a \Rightarrow a = \frac{3 d}{2} . . . (i) And, a_{7} - 2 a_{3} = 1 \Rightarrow a + (7 - 1) d - 2 [a + (3 - 1) d] = 1 \Rightarrow a + 6 d - 2 (a + 2 d) = 1 \Rightarrow a + 6 d - 2 a - 4 d = 1 \Rightarrow - a + 2 d = 1 \Rightarrow - \frac{3 d}{2} + 2 d = 1 [From (i)] \Rightarrow \frac{- 3 d + 4 d}{2} = 1 \Rightarrow \frac{d}{2} = 1 \Rightarrow d = 2$
$Putting the value in (i), we get : a = \frac{3 \times 2}{2} \Rightarrow a = 3$

Page No 19.12:

Question 17:

Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.

Answer:

Given:
$a_{6} = 12 \Rightarrow a + (6 - 1) d = 12 \Rightarrow a + 5 d = 12 . . . (i) a_{8} = 22 \Rightarrow a + (8 - 1) d = 22 \Rightarrow a + 7 d = 22 . . . (ii) Solving (i) and (ii), we get : 2 d = 10 \Rightarrow d = 5 Putting the value of d in (i), we get : a + 5 \times 5 = 12 \Rightarrow a = 12 - 25 = - 13 ∴ a_{2} = a + (2 - 1) d = a + d = - 13 + 5 = - 8 Also, a_{n} = a + (n - 1) d = - 13 + (n - 1) 5 = - 13 + 5 n - 5 = 5 n - 18$

Page No 19.12:

Question 18:

How many numbers of two digit are divisible by 3?

Answer:

The two digit numbers that are divisible by 3 are:
12, 15, 18...96, 99
This is an A.P. whose first term is 12 and the common difference is 3.
$We have : a_{n} = 99 \Rightarrow 12 + (n - 1) 3 = 99 \Rightarrow (n - 1) 3 = 87 \Rightarrow (n - 1) = 29 \Rightarrow n = 30$
Thus, there are 30 such terms.

Page No 19.12:

Question 19:

An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.

Answer:

Given:
$a = 7, n = 60, l = 125$
$l = a + (n - 1) d \Rightarrow 125 = 7 + (60 - 1) d \Rightarrow 125 = 7 + 59 d \Rightarrow 118 = 59 d \Rightarrow 2 = d$

$a_{32} = a + (32 - 1) d = a + 31 d = 7 + 31 \times 2 = 7 + 62 = 69$

Page No 19.12:

Question 20:

The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.

Answer:

Let a be the first term and d be the common difference. Then,
$a_{4} + a_{8} = 24 \Rightarrow a + (4 - 1) d + a + (8 - 1) d = 24 \Rightarrow a + 3 d + a + 7 d = 24 \Rightarrow 2 a + 10 d = 24 \Rightarrow a + 5 d = 12 . . . (i) Also, a_{6} + a_{10} = 34 \Rightarrow a + (6 - 1) d + a + (10 - 1) d = 34 \Rightarrow a + 5 d + a + 9 d = 34 \Rightarrow 2 a + 14 d = 34 \Rightarrow a + 7 d = 17 . . . (ii) Solving (i) and (ii), we get : 2 d = 5 \Rightarrow d = \frac{5}{2} Substituing the value in (i), we get : a + 5 (\frac{5}{2}) = 12 \Rightarrow a + \frac{25}{2} = 12 \Rightarrow a = \frac{- 1}{2}$

Page No 19.12:

Question 21:

How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?

Answer:

A number N is divided by 7 leaves a remainder 4.

∴ N = 7q + 4

N can take values 4, 11, 18, ..... 998

Now,
4, 11, 18, ..... 998 are in arithmetic progression.
First term a = 4
common difference d = 7
last term l = 998

We know that,
l = a + (n − 1)d
⇒ 998 = 4 + (n − 1)7
⇒ 998 = 4 + 7n − 7
⇒ 998 = 7n − 3
⇒ 1001 = 7n
⇒ $n = \frac{1001}{7}$
⇒ n = 143

Hence, 143 numbers are there between 1 and 1000 which when divided by 7 leave remainder 4.

Page No 19.12:

Question 22:

The first and the last terms of an A.P. are a and l respectively. Show that the sum of nth term from the beginning and nth term from the end is a + l.

Answer:

Given:
First term = a
Last term = l
nth term from the beginning = $a + (n - 1) d$ , where d is the common difference.
nth term from the end = $l + (n - 1) (- d) = l - d n + d$
Their sum = $a + (n - 1) d + l - d n + d$
$= a + n d - d + l - n d + d = a + l$
Hence, proved.

Page No 19.12:

Question 23:

If < a_n > is an A.P. such that $\frac{a_{4}}{a_{7}} = \frac{2}{3}, find \frac{a_{6}}{a_{8}}$ .

Answer:

Given:
< a_n > is an A.P.
$\frac{a_{4}}{a_{7}} = \frac{2}{3} \Rightarrow \frac{a + (4 - 1) d}{a + (7 - 1) d} = \frac{2}{3} \Rightarrow \frac{a + 3 d}{a + 6 d} = \frac{2}{3} \Rightarrow 3 (a + 3 d) = 2 (a + 6 d) \Rightarrow 3 a + 9 d = 2 a + 12 d \Rightarrow a = 3 d . . . . (i)$

$∴ \frac{a_{6}}{a_{8}} = \frac{a + (6 - 1) d}{a + (8 - 1) d} \Rightarrow \frac{a_{6}}{a_{8}} = \frac{a + 5 d}{a + 7 d} \Rightarrow \frac{a_{6}}{a_{8}} = \frac{3 d + 5 d}{3 d + 7 d} (From (i)) \Rightarrow \frac{a_{6}}{a_{8}} = \frac{8 d}{10 d} \Rightarrow \frac{a_{6}}{a_{8}} = \frac{4 d}{5 d} = \frac{4}{5}$

Page No 19.12:

Question 24:

$If θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n} are in AP, whose common difference is d, then show that \sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + . . . + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d} [NCERT EXEMPLAR]$

Answer:

$As, θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n} are in AP So, d = θ_{2} - θ_{1} = θ_{3} - θ_{2} = . . . = θ_{n} - θ_{n - 1} . . . . . (i) Now, LHS = \sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + . . . + \sec θ_{n - 1} \sec θ_{n} = \frac{1}{\cos θ_{1} \cos θ_{2}} + \frac{1}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{1}{\cos θ_{n - 1} \cos θ_{n}} = \frac{1}{\sin d} [\frac{\sin d}{\cos θ_{1} \cos θ_{2}} + \frac{\sin d}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin d}{\cos θ_{n - 1} \cos θ_{n}}] = \frac{1}{\sin d} [\frac{\sin (θ_{2} - θ_{1})}{\cos θ_{1} \cos θ_{2}} + \frac{\sin (θ_{3} - θ_{2})}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin (θ_{n} - θ_{n - 1})}{\cos θ_{n - 1} \cos θ_{n}}] [Using (i)] = \frac{1}{\sin d} [\frac{\sin θ_{2} \cos θ_{1} - \cos θ_{2} \sin θ_{1}}{c o s θ_{1} c o s θ_{2}} + \frac{\sin θ_{3} \cos θ_{2} - \cos θ_{3} \sin θ_{2}}{c o s θ_{2} c o s θ_{3}} + . . . + \frac{\sin θ_{n} \cos θ_{n - 1} - \cos θ_{n} \sin θ_{n - 1}}{c o s θ_{n - 1} c o s θ_{n}}] = \frac{1}{\sin d} [\frac{\sin θ_{2} \cos θ_{1}}{\cos θ_{1} \cos θ_{2}} - \frac{\cos θ_{2} \sin θ_{1}}{\cos θ_{1} \cos θ_{2}} + \frac{\sin θ_{3} \cos θ_{2}}{\cos θ_{2} \cos θ_{3}} - \frac{\cos θ_{3} \sin θ_{2}}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin θ_{n} \cos θ_{n - 1}}{\cos θ_{n - 1} \cos θ_{n}} - \frac{\cos θ_{n} \sin θ_{n - 1}}{\cos θ_{n - 1} \cos θ_{n}}] = \frac{1}{\sin d} [\tan θ_{2} - \tan θ_{1} + \tan θ_{3} - \tan θ_{2} + . . . + \tan θ_{n} - \tan θ_{n - 1}] = \frac{1}{\sin d} [- \tan θ_{1} + \tan θ_{n}] = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d} = RHS$

Page No 19.15:

Question 1:

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Answer:

$Let the three terms of the A . P . be a - d, a, a + d . Then, we have : a - d + a + a + d = 21 \Rightarrow 3 a = 21 \Rightarrow a = 7 . . . . (i) Also, (a - d) (a + d) - a = 6 \Rightarrow a^{2} - d^{2} - a = 6 \Rightarrow 49 - d^{2} - 7 = 6 \Rightarrow 36 = d^{2} \Rightarrow \pm 6 = d When d = 6, a = 7, we get : 1, 7, 13 When d = - 6, a = 7, we get : 13, 7, 1$

Page No 19.15:

Question 2:

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

Answer:

$Let the three numbers be a - d, a, a + d . Their sum = 27 \Rightarrow a - d + a + a + d = 27 \Rightarrow 3 a = 27 \Rightarrow a = 9 . . . (i) Product = (a - d) a (a + d) = 648 \Rightarrow a (a^{2} - d^{2}) = 648 \Rightarrow 9 (81 - d^{2}) = 648 \Rightarrow (81 - d^{2}) = 72 \Rightarrow d^{2} = 9 \Rightarrow d = \pm 3 When a = 9, d = 3, we have : 6, 9, 12 When a = 9, d = - 3, we have : 12, 9, 6$

Page No 19.15:

Question 3:

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Answer:

$Let the four numbers be a - 3 d, a - d, a + d, a + 3 d . Sum = 50 \Rightarrow a - 3 d + a - d + a + d + a + 3 d = 50 \Rightarrow 4 a = 50 \Rightarrow a = \frac{25}{2} . . . (i) Also, a + 3 d = 4 (a - 3 d) \Rightarrow a + 3 d = 4 a - 12 d \Rightarrow 3 a = 15 d \Rightarrow a = 5 d \Rightarrow \frac{25}{2 \times 5} = d (Using (i)) \Rightarrow \frac{5}{2} = d So, the terms are as follows : (\frac{25}{2} - 3 \times \frac{5}{2}), (\frac{25}{2} - \frac{5}{2}), (\frac{25}{2} + \frac{5}{2}), (\frac{25}{2} + 3 \times \frac{5}{2}) = 5, 10, 15, 20$

Page No 19.15:

Question 4:

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Answer:

$Let the numbers be (a - d), a, (a + d) . Sum = a - d + a + a + d = 12 \Rightarrow 3 a = 12 \Rightarrow a = 4 Also, (a - d)^{3} + a^{3} + (a + d)^{3} = 288 \Rightarrow a^{3} - d^{3} - 3 a^{2} d + 3 a d^{2} + a^{3} + a^{3} + d^{3} + 3 a^{2} d + 3 a d^{2} = 288 \Rightarrow 3 a^{3} + 6 a d^{2} = 288 \Rightarrow 3 {(4)}^{3} + 6 \times 4 \times d^{2} = 288 \Rightarrow 192 + 24 d^{2} = 288 \Rightarrow 24 d^{2} = 96 \Rightarrow d^{2} = 4 \Rightarrow d = \pm 2 When a = 4, d = 2, the numbers are 2, 4, 6 . When a = 4, d = - 2, the numbers are 6, 4, 2 .$

Page No 19.15:

Question 5:

If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Answer:

$Let the three numbers be (a - d), a, (a + d) . Sum = 24 \Rightarrow (a - d) + a + (a + d) = 24 \Rightarrow 3 a = 24 \Rightarrow a = 8 . . . (i) Product = a (a - d) (a + d) = 440 \Rightarrow a (a^{2} - d^{2}) = 440 \Rightarrow 8 (64 - d^{2}) = 440 (Form (i)) \Rightarrow (64 - d^{2}) = 55 \Rightarrow d^{2} = 9 \Rightarrow d = \pm 3$

$With a = 8, d = 3, we have : 5, 8, 11 With a = 8, d = - 3, we have : 11, 8, 5$

Page No 19.15:

Question 6:

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

Answer:

Let the angles be $(A) °, (A + d) °, (A + 2 d) °, (A + 3 d) °$ .
Here, d = 10
So, $(A) °, (A + 10) °, (A + 20) °, (A + 30) °$ are the angles of a quadrilateral whose sum is 360^o.

$∴ (A) °, (A + 10) °, (A + 20) °, (A + 30) ° = 360 ° \Rightarrow 4 A = 360 - 60 \Rightarrow A = \frac{300}{4} = 75 °$

$The angles are as follows : 75 °, (75 + 10) °, (75 + 20) °, (75 + 30) °, i . e . 75 °, 85 °, 95 °, 105 °$

Page No 19.30:

Question 1:

Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, ... to 10 terms
(ii) 1, 3, 5, 7, ... to 12 terms
(iii) 3, 9/2, 6, 15/2, ... to 25 terms
(iv) 41, 36, 31, ... to 12 terms
(v) a + b, a − b, a − 3b, ... to 22 terms
(vi) (x − y)², (x² + y²), (x + y)², ... to n terms
(vii) $\frac{x - y}{x + y}, \frac{3 x - 2 y}{x + y}, \frac{5 x - 3 y}{x + y}$ , ... to n terms

Answer:

(i) 50, 46, 42 ... to 10 terms
$We have : a = 50, d = (46 - 50) = - 4 n = 10 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{10}{2} [2 \times 50 + (10 - 1) (- 4)] = 5 [100 - 36] = 320$

(ii) 1, 3, 5, 7 ... to 12 terms
$We have : a = 1, d = (3 - 1) = 2 n = 12 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{12}{2} [2 \times 1 + (12 - 1) (2)] = 6 [24] = 144$

(iii) 3, 9/2, 6, 15/2 ... to 25 terms
$We have : a = 3, d = (9 / 2 - 3) = 3 / 2 n = 25 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{25}{2} [2 \times 3 + (25 - 1) (3 / 2)] = \frac{25}{2} \times 42 = 525$

(iv) 41, 36, 31 ... to 12 terms
$We have : a = 41, d = (36 - 41) = - 5 n = 12 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{12}{2} [2 \times 41 + (12 - 1) (- 5)] = 6 \times 27 = 162$

(v) a + b, a − b, a − 3b ... to 22 terms
$We have : First term = a + b, d = (a - b - a - b) = - 2 b n = 22 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{22}{2} [2 \times (a + b) + (22 - 1) (- 2 b)] = 11 [2 a - 40 b] = 22 a - 440 b$

(vi) (x − y)², (x² + y²), (x + y)² ... to n terms
$We have : a = {(x - y)}^{2}, d = (x^{2} + y^{2} - {(x - y)}^{2}) = 2 x y S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [2 {(x - y)}^{2} + (n - 1) (2 x y)] = \frac{n}{2} \times 2 [{(x - y)}^{2} + (n - 1) (x y)] = n [{(x - y)}^{2} + (n - 1) (x y)]$

(vii) $\frac{x - y}{x + y}, \frac{3 x - 2 y}{x + y}, \frac{5 x - 3 y}{x + y}$ ... to n terms
$We have : a = \frac{x - y}{x + y}, d = (\frac{3 x - 2 y}{x + y} - \frac{x - y}{x + y}) = (\frac{2 x - y}{x + y}) S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [2 (\frac{x - y}{x + y}) + (n - 1) (\frac{2 x - y}{x + y})] = \frac{n}{2 (x + y)} [(2 x - 2 y) + (2 x - y) (n - 1)] = \frac{n}{2 (x + y)} [2 x - 2 y - 2 x + y + n (2 x - y)] = \frac{n}{2 (x + y)} [n (2 x - y) - y]$

Page No 19.30:

Question 2:

Find the sum of the following series:
(i) 2 + 5 + 8 + ... + 182
(ii) 101 + 99 + 97 + ... + 47
(iii) (a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]

Answer:

(i) 2 + 5 + 8 + ... + 182
Here, the series is an A.P. where we have the following:
$a = 2 d = (5 - 2) = 3 a_{n} = 182 \Rightarrow 2 + (n - 1) (3) = 182 \Rightarrow 2 + 3 n - 3 = 182 \Rightarrow 3 n - 1 = 182 \Rightarrow 3 n = 183 \Rightarrow n = 61 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{61} = \frac{61}{2} [2 \times 2 + (61 - 1) \times 3] = \frac{61}{2} [2 \times 2 + 60 \times 3] = 5612$

(ii) 101 + 99 + 97 + ... + 47
Here, the series is an A.P. where we have the following:
$a = 101 d = (99 - 101) = - 2 a_{n} = 47 \Rightarrow 101 + (n - 1) (- 2) = 47 \Rightarrow 101 - 2 n + 2 = 47 \Rightarrow 2 n - 2 = 54 \Rightarrow 2 n = 56 \Rightarrow n = 28 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{28} = \frac{28}{2} [2 \times 101 + (28 - 1) \times (- 2)] = \frac{28}{2} [2 \times 101 + 27 \times (- 2)] = 2072$

(iii) (a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]
Here, the series is an A.P. where we have the following:
$a = (a - b)^{2} d = (a^{2} + b^{2} - (a - b)^{2}) = 2 a b a_{n} = {[(a + b)}^{2} + 6 a b] \Rightarrow (a - b)^{2} + (n - 1) (2 a b) = {[(a + b)}^{2} + 6 a b] \Rightarrow a^{2} + b^{2} - 2 a b + 2 a b n - 2 a b = [a^{2} + b^{2} + 2 a b + 6 a b] \Rightarrow a^{2} + b^{2} - 4 a b + 2 a b n = a^{2} + b^{2} + 8 a b \Rightarrow 2 a b n = 12 a b \Rightarrow n = 6 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{6} = \frac{6}{2} [2 (a - b)^{2} + (6 - 1) 2 a b] = 3 [2 (a^{2} + b^{2} - 2 a b) + 10 a b] = 3 [2 a^{2} + 2 b^{2} - 4 a b + 10 a b] = 3 [2 a^{2} + 2 b^{2} + 6 a b] = 6 [a^{2} + b^{2} + 3 a b]$

Page No 19.30:

Question 3:

Find the sum of first n natural numbers.

Answer:

The first n natural numbers are:
1, 2, 3, 4...
a = 1, d = 1, Total terms = n

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{n} = \frac{n}{2} [2 \times 1 + (n - 1) 1] \Rightarrow S_{n} = \frac{n}{2} [2 + (n - 1) 1] \Rightarrow S_{n} = \frac{n}{2} [n + 1]$

Page No 19.30:

Question 4:

Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5.

Answer:

We have to find the sum of all the natural numbers that are divisible by 2 or 5
Required Sum = Sum of the natural numbers between 1 and 100 that are divisible by 2 + Sum of the natural numbers between 1 and 100 that are divisible by 5
                      − Sum of the natural numbers between 1 and 100 that are divisible by 2 and 5, i.e by 10

         $= (2 + 4 + 6 + 8 + . . . + 98) + (5 + 10 + 15 + . . . + 95) - (10 + 20 + 30 + . . . + 90) = \frac{50}{2} (2 + 98) + \frac{20}{2} (5 + 95) - \frac{10}{2} (10 + 90) = 2500 + 1000 - 500 = 3000$

Page No 19.30:

Question 5:

Find the sum of first n odd natural numbers.

Answer:

The first n odd natural numbers are:
1, 3, 5, 7, 9...
a = 1, d = 2, Total terms = n

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{n} = \frac{n}{2} [2 \times 1 + (n - 1) 2] \Rightarrow S_{n} = \frac{n}{2} [2 + (n - 1) 2] \Rightarrow S_{n} = \frac{n}{2} [2 n] \Rightarrow S_{n} = n^{2}$

Page No 19.30:

Question 6:

Find the sum of all odd numbers between 100 and 200.

Answer:

All the odd numbers between 100 and 200 are:
101, 103...199
Here, we have:
$a = 101 d = 2 a_{n} = 199 \Rightarrow 101 + (n - 1) \times 2 = 199 \Rightarrow 2 n - 2 = 98 \Rightarrow 2 n = 100 \Rightarrow n = 50 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{50} = \frac{50}{2} [2 \times 101 + (50 - 1) 2] \Rightarrow S_{50} = 25 [202 + 98]$
$\Rightarrow S_{50} = 7500$

Page No 19.30:

Question 7:

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

Answer:

The odd integers between 1 and 1000 that are divisible by 3 are:
3, 9, 15, 21...999
Here, we have:
$a = 3, d = 6 a_{n} = 999 \Rightarrow 3 + (n - 1) 6 = 999 \Rightarrow 3 + 6 n - 6 = 999 \Rightarrow 6 n = 1002 \Rightarrow n = 167 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{167} = \frac{167}{2} [2 \times 3 + (167 - 1) 6] \Rightarrow S_{167} = \frac{167}{2} [1002] = 83667 Hence, proved .$

Page No 19.30:

Question 8:

Find the sum of all integers between 84 and 719, which are multiples of 5.

Answer:

The integers between 84 and 719, which are multiples of 5 are:
85, 90...715
Here, we have:
$a = 85 d = 5 a_{n} = 715 \Rightarrow 85 + (n - 1) 5 = 715 \Rightarrow 5 n - 5 = 630 \Rightarrow n = 127 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{127} = \frac{127}{2} [2 \times 85 + (127 - 1) 5] \Rightarrow S_{127} = \frac{127}{2} [800] = 50800$

Page No 19.31:

Question 9:

Find the sum of all integers between 50 and 500 which are divisible by 7.

Answer:

The integers between 50 and 500 that are divisible by 7 are:
56, 63...497
Here, we have:
$a = 56 d = 7 a_{n} = 497 \Rightarrow 56 + (n - 1) 7 = 497 \Rightarrow 7 n - 7 = 441 \Rightarrow 7 n = 448 \Rightarrow n = 64 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{64} = \frac{64}{2} [2 \times 56 + (64 - 1) 7] \Rightarrow S_{64} = 32 [2 \times 56 + 63 \times 7] = 17696$

Page No 19.31:

Question 10:

Find the sum of all even integers between 101 and 999.

Answer:

The even integers between 101 and 999 are:
102, 104...998
Here, we have:
$a = 102 d = 2 a_{n} = 998 \Rightarrow 102 + (n - 1) 2 = 998 \Rightarrow 2 n - 2 = 896 \Rightarrow 2 n = 898 \Rightarrow n = 449 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{449} = \frac{449}{2} [2 \times 102 + (449 - 1) \times 2] \Rightarrow S_{449} = \frac{449}{2} [1100] \Rightarrow S_{449} = 246950$

Page No 19.31:

Question 11:

Find the sum of all integers between 100 and 550, which are divisible by 9.

Answer:

The integers between 100 and 550 that are divisible by 9 are:
108, 117...549
Here, we have:
$a = 108 d = 9 a_{n} = 549 \Rightarrow 108 + (n - 1) (9) = 549 \Rightarrow 9 n - 9 = 441 \Rightarrow 9 n = 450 \Rightarrow n = 50 S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{50} = \frac{50}{2} [2 \times 108 + (50 - 1) \times 9] \Rightarrow S_{50} = 25 (657) = 16425$

Page No 19.31:

Question 12:

Find the sum of the series:
3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3n terms.

Answer:

The given sequence i.e., 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 +..... to 3n terms.

can be rewritten as 3 + 6 + 9 + .... to n terms + 5 + 9 + 13 + .... to n terms + 7 + 12 + 17 + .... to n terms

Clearly, all these sequence forms an A.P. having n terms with first terms 3, 5, 7 and common difference 3, 4, 5

Hence, required sum = $\frac{n}{2} [2 \times 3 + (n - 1) 3] + \frac{n}{2} [2 \times 5 + (n - 1) 4] + \frac{n}{2} [2 \times 7 + (n - 1) 5]$

$= \frac{n}{2} [(6 + 3 n - 3) + (10 + 4 n - 4) + (14 + 5 n - 5)] = \frac{n}{2} [12 n + 18] = 3 n (2 n + 3)$

Page No 19.31:

Question 13:

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.

Answer:

The sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 are:
103, 119...791
Here, we have:
a = 103
d = 16
$a_{n} = 791$
$We know : a_{n} = a + (n - 1) d \Rightarrow 791 = 103 + (n - 1) \times 16 \Rightarrow 688 = 16 n - 16 \Rightarrow 704 = 16 n \Rightarrow 44 = n Also, S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{44} = \frac{44}{2} [2 \times 103 + (44 - 1) \times 16] \Rightarrow S_{44} = 22 [206 + 688] \Rightarrow S_{44} = 22 \times 894 = 19668$

Page No 19.31:

Question 14:

Solve: (i) 25 + 22 + 19 + 16 + ... + x = 115
(ii) 1 + 4 + 7 + 10 + ... + x = 590.

Answer:

(i) 25 + 22 + 19 + 16 + ... + x = 115
Here, a = 25, d = $-$ 3, $S_{n} = 115$
$We know : S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow 115 = \frac{n}{2} [2 \times 25 + (n - 1) \times (- 3)] \Rightarrow 115 \times 2 = n [50 - 3 n + 3] \Rightarrow 230 = n (53 - 3 n) \Rightarrow 230 = 53 n - 3 n^{2} \Rightarrow 3 n^{2} - 53 n + 230 = 0 By quadratic formula : n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} Substituting a = 3, b = - 53 and c = 230, we get : n = \frac{53 \pm \sqrt{{(53)}^{2} - 4 \times 3 \times 230}}{2 \times 3} = \frac{46}{6}, 10 \Rightarrow n = 10, a s n \neq \frac{46}{6} ∴ a_{n} = a + (n - 1) d \Rightarrow x = 25 + (10 - 1) (- 3) \Rightarrow x = 25 - 27 = - 2$

(ii) 1 + 4 + 7 + 10 + ... + x = 590
Here, a = 1, d = 3,
$We know : S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow 590 = \frac{n}{2} [2 \times 1 + (n - 1) \times (3)] \Rightarrow 590 \times 2 = n [2 + 3 n - 3] \Rightarrow 1180 = n (3 n - 1) \Rightarrow 1180 = 3 n^{2} - n \Rightarrow 3 n^{2} - n - 1180 = 0 By quadratic formula : n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} Substituting a = 3, b = - 1 and c = - 1180, we get : \Rightarrow n = \frac{1 \pm \sqrt{{(1)}^{2} + 4 \times 3 \times 1180}}{2 \times 3} = \frac{- 118}{6}, 20 \Rightarrow n = 20, a s n \neq \frac{- 118}{6} ∴ a_{n} = x = a + (n - 1) d \Rightarrow x = 1 + (20 - 1) (3) \Rightarrow x = 1 + 60 - 3 = 58$

Page No 19.31:

Question 15:

Find the rth term of an A.P., the sum of whose first n terms is 3n² + 2n. [NCERT EXEMPLAR]

Answer:

$Let a and d be the first term and the common difference of the given A . P ., respectively As, S_{n} = 3 n^{2} + 2 n So, a = S_{1} = 3 \times 1^{2} + 2 \times 1 = 3 + 2 = 5 and S_{2} = 3 \times 2^{2} + 2 \times 2 = 12 + 4 = 16 \Rightarrow a + a_{2} = 16 \Rightarrow a + a + d = 16 \Rightarrow 2 a + d = 16 \Rightarrow 2 \times 5 + d = 16 \Rightarrow d = 16 - 10 \Rightarrow d = 6 Now, a_{r} = a + (r - 1) d = 5 + (r - 1) \times 6 = 5 + 6 r - 6 ∴ a_{r} = 6 r - 1$

Page No 19.31:

Question 16:

How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

Answer:

$We have : a = - 14 and S_{n} = 40 . . . (i) a_{5} = 2 \Rightarrow a + (5 - 1) d = 2 \Rightarrow - 14 + 4 d = 2 \Rightarrow 4 d = 16 \Rightarrow d = 4 . . . (ii) Also, S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow 40 = \frac{n}{2} [2 (- 14) + (n - 1) \times 4] (From (i) and (ii)) \Rightarrow 80 = n [- 28 + 4 n - 4] \Rightarrow 80 = 4 n^{2} - 32 n \Rightarrow n^{2} - 8 n - 20 = 0 \Rightarrow (n - 10) (n + 2) = 0 \Rightarrow n = 10, - 2 But, n cannot be negative . ∴ n = 10$

Page No 19.31:

Question 17:

The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.

Answer:

$We have : S_{7} = 10 \Rightarrow \frac{7}{2} [2 a + (7 - 1) d] = 10 \Rightarrow \frac{7}{2} [2 a + 6 d] = 10 \Rightarrow a + 3 d = \frac{10}{7} . . . (i) Also, the sum of the next seven terms = S_{14} - S_{7} = 17 \Rightarrow \frac{14}{2} [2 a + (14 - 1) d] - \frac{7}{2} [2 a + (7 - 1) d] = 17 \Rightarrow 7 [2 a + 13 d] - \frac{7}{2} [2 a + 6 d] = 17 \Rightarrow 14 a + 91 d - 7 a - 21 d = 17 \Rightarrow 7 a + 70 d = 17 \Rightarrow a + 10 d = \frac{17}{7} . . . (ii) From (i) and (ii), we get : \frac{10}{7} - 3 d = \frac{17}{7} - 10 d \Rightarrow 7 d = 1 \Rightarrow d = \frac{1}{7} Putting the value in (i), we get : a + 3 d = \frac{10}{7} \Rightarrow a + \frac{3}{7} = \frac{10}{7} \Rightarrow a = 1 ∴ a = 1, d = \frac{1}{7}$
The progression thus formed is $1, \frac{8}{7}, \frac{9}{7}, \frac{10}{7} . . .$

Page No 19.31:

Question 18:

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

Answer:

$Given : a_{3} = 7, a_{7} - 3 a_{3} = 2 We have : a_{3} = 7 \Rightarrow a + (3 - 1) d = 7 \Rightarrow a + 2 d = 7 . . . (i) Also, a_{7} - 3 a_{3} = 2 \Rightarrow a_{7} - 21 = 2 (Given) \Rightarrow a + (7 - 1) d = 23 \Rightarrow a + 6 d = 23 . . . (ii) From (i) and (ii), we get : 4 d = 16 \Rightarrow d = 4 Putting the value in (i), we get : a + 2 (4) = 7 \Rightarrow a = - 1 ∴ S_{20} = \frac{20}{2} [2 (- 1) + (20 - 1) (4)] \Rightarrow S_{20} = 10 [- 2 + 76] \Rightarrow S_{20} = 10 [74] = 740 ∴ a = - 1, d = 4, S_{20} = 740$

Page No 19.31:

Question 19:

The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

Answer:

$Given : a = 2, l = 50, S_{n} = 442 Now, S_{n} = 442 \Rightarrow \frac{n}{2} [a + l] = 442 \Rightarrow \frac{n}{2} [2 + 50] = 442 \Rightarrow n = 17 ∴ a_{17} = 50 \Rightarrow a + (17 - 1) \times d = 50 \Rightarrow 2 + 16 d = 50 \Rightarrow d = 3$

Page No 19.31:

Question 20:

The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by $10 \frac{1}{2}$ , find the number of terms and the series.

Answer:

Let total number of terms be 2n.
According to question, we have:

$a_{1} + a_{3} + . . . + a_{2 n - 1} = 24 . . . (1) a_{2} + a_{4} + . . . + a_{2 n} = 30 . . . (2) Subtracting (1) from (2), we get : (d + d + . . . + upto n terms) = 6 \Rightarrow n d = 6 . . . (3) Given : a_{2 n} = a_{1} + \frac{21}{2} \Rightarrow a_{2 n} - a_{1} = \frac{21}{2} \Rightarrow a + (2 n - 1) d - a = \frac{21}{2} [∵ a_{2 n} = a + (2 n - 1) d, a_{1} = a] \Rightarrow 2 n d - d = \frac{21}{2} \Rightarrow 2 \times 6 - d = \frac{21}{2} (From (3)) \Rightarrow d = \frac{3}{2} Putting the value in (3), we get : n = 4 \Rightarrow 2 n = 8 Thus, there are 8 terms in the progression .$

$To find the value of the first term : a_{2} + a_{4} + . . . + a_{2 n} = 30 \Rightarrow (a + d) + (a + 3 d) + . . . + [a + (2 n - 1) d] = 30 \Rightarrow \frac{n}{2} [(a + d) + a + (2 n - 1) d] = 30 Putting n = 4 and d = \frac{3}{2}, we get : a = \frac{3}{2} So, the series will be 1 \frac{1}{2}, 3, 4 \frac{1}{2} . . .$

Page No 19.31:

Question 21:

If S_n = n²p and S_m = m²p, m ≠ n, in an A.P., prove that S_p = p³.

Answer:

$S_{n} = n^{2} p \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = n^{2} p \Rightarrow 2 n p = 2 a + (n - 1) d . . . (i) S_{m} = m^{2} p \Rightarrow \frac{m}{2} [2 a + (m - 1) d] = m^{2} p \Rightarrow 2 m p = 2 a + (m - 1) d . . . (i i) Subtracting (ii) from (i), we get : 2 p (n - m) = (n - m) d \Rightarrow 2 p = d . . . (i i i) Substituing the value in (i), we get : n d = 2 a + (n - 1) d \Rightarrow n d - n d + d = 2 a \Rightarrow a = \frac{d}{2} = p [from (iii)] . . . (iv) ∴ S_{p} = \frac{p}{2} [2 a + (p - 1) d] \Rightarrow S_{p} = \frac{p}{2} [2 p + (p - 1) 2 p] \Rightarrow S_{p} = \frac{p}{2} [2 p + 2 p^{2} - 2 p] \Rightarrow S_{p} = \frac{p}{2} [2 p^{2}] \Rightarrow S_{p} = p^{3}$

Page No 19.31:

Question 22:

If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Answer:

Let a be the first term and d be the common difference.
$a_{12} = - 13 \Rightarrow a + (12 - 1) d = - 13 \Rightarrow a + 11 d = - 13 . . . (i) Also, S_{4} = 24 \Rightarrow \frac{4}{2} [2 a + (4 - 1) d] = 24 \Rightarrow 2 (2 a + 3 d) = 24 \Rightarrow 2 a + 3 d = 12 . . . (i i) From (i) and (ii), we get : 19 d = - 38 \Rightarrow d = - 2 Putting the value of d in (i), we get : a + 11 (- 2) = - 13 \Rightarrow a = 9 S_{10} = \frac{10}{2} [2 \times 9 + (10 - 1) (- 2)] \Rightarrow S_{10} = 5 [18 + 9 (- 2)] = 0$

Page No 19.31:

Question 23:

If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?

Answer:

$We have : a_{5} = 30 \Rightarrow a + (5 - 1) d = 30 \Rightarrow a + 4 d = 30 . . . (i) Also, a_{12} = 65 \Rightarrow a + (12 - 1) d = 65 \Rightarrow a + 11 d = 65 . . . . . (ii) Solving (i) and (ii), we get : 7 d = 35 \Rightarrow d = 5 Putting the value of d in (i), we get : a + 4 \times 5 = 30 \Rightarrow a = 10 ∴ S_{20} = \frac{20}{2} [2 \times 10 + (20 - 1) \times 5] \Rightarrow S_{20} = 10 [2 \times 10 + (20 - 1) \times 5] \Rightarrow S_{20} = 1150$

Page No 19.31:

Question 24:

Find the sum of n terms of the A.P. whose kth terms is 5k + 1.

Answer:

$Given : a_{k} = 5 k + 1 For k = 1, a_{1} = 5 \times 1 + 1 = 6 For k = 2, a_{2} = 5 \times 2 + 1 = 11 For k = n, a_{n} = 5 n + 1 ∴ S_{n} = \frac{n}{2} [a + a_{n}] \Rightarrow S_{n} = \frac{n}{2} [6 + 5 n + 1] = \frac{n}{2} (5 n + 7)$

Page No 19.31:

Question 25:

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer:

The two-digit numbers which when divided by 4 yield 1 as remainder are 13, 17....97.

$∴ a = 13, d = 4, a_{n} = 97 ∴ a_{n} = a + (n - 1) d \Rightarrow 97 = 13 + (n - 1) 4 \Rightarrow 84 = 4 n - 4 \Rightarrow 88 = 4 n \Rightarrow 22 = n . . . (1) Also, S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{22} = \frac{22}{2} [2 \times 13 + (22 - 1) \times 4] (From (1)) \Rightarrow S_{22} = 11 [110] = 1210$

Page No 19.31:

Question 26:

If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.

Answer:

The given A.P. is 25,.22,.19.....
Here, a = 25, d = $22 - 25 = - 3$
$S_{n} = 116 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 116 \Rightarrow n [2 \times 25 + (n - 1) (- 3)] = 232 \Rightarrow 50 n - 3 n^{2} + 3 n = 232 \Rightarrow 3 n^{2} - 53 n + 232 = 0$

$\Rightarrow 3 n^{2} - 29 n - 24 n + 232 = 0 \Rightarrow n (3 n - 29) - 8 (3 n - 29) = 0 \Rightarrow (3 n - 29) (n - 8) = 0 \Rightarrow n = \frac{29}{3} o r 8 Since n cannot be a fraction, n = 8 . Thus, the last term : a_{n} = a + (n - 1) d \Rightarrow a_{8} = 25 + (8 - 1) \times (- 3) \Rightarrow a_{8} = 4$

Page No 19.31:

Question 27:

Find the sum of odd integers from 1 to 2001.

Answer:

$The odd integers from 1 to 2001 are 1, 3, 5 . . . . . 2001 . It is an AP with a = 1 and d = 2 . a_{n} = 2001 \Rightarrow 1 + (n - 1) 2 = 2001 \Rightarrow 2 n - 2 = 2000 \Rightarrow 2 n = 2002 \Rightarrow n = 1001 Also, S_{1001} = \frac{1001}{2} [2 \times 1 + (1001 - 1) 2] \Rightarrow S_{1001} = \frac{1001}{2} [2 \times 1 + (1000) 2] \Rightarrow S_{1001} = \frac{1001}{2} \times 2002 = 1002001$

Page No 19.31:

Question 28:

How many terms of the A.P. −6, $- \frac{11}{2}$ , −5, ... are needed to give the sum −25?

Answer:

$Given : An A . P . with a = - 6 and d = - \frac{11}{2} - (- 6) = \frac{1}{2} S_{n} = - 25 ∴ - 25 = \frac{n}{2} [2 \times (- 6) + (n - 1) \frac{1}{2}] \Rightarrow - 25 = \frac{n}{2} [- 12 + \frac{n}{2} - \frac{1}{2}] \Rightarrow - 50 = n [\frac{n}{2} - \frac{25}{2}] \Rightarrow - 100 = n (n - 25) \Rightarrow n^{2} - 25 n + 100 = 0 \Rightarrow (n - 20) (n - 5) = 0 \Rightarrow n = 20 or n = 5$

Page No 19.31:

Question 29:

In an A.P. the first term is 2 and the sum of the first five terms is one fourth of the next five terms. Show that 20th term is −112.

Answer:

$Given : a = 2, S_{5} = \frac{1}{4} (S_{10} - S_{5}) We have : S_{5} = \frac{5}{2} [2 \times 2 + (5 - 1) d] \Rightarrow S_{5} = 5 [2 + 2 d] . . . . (i) Also, S_{10} = \frac{10}{2} [2 \times 2 + (10 - 1) d] \Rightarrow S_{10} = 5 [4 + 9 d] . . . . . (i i) ∵ S_{5} = \frac{1}{4} (S_{10} - S_{5}) From (i) and (ii), we have : \Rightarrow 5 [2 + 2 d] = \frac{1}{4} [5 (4 + 9 d) - 5 (2 + 2 d)] \Rightarrow 8 + 8 d = 4 + 9 d - 2 - 2 d \Rightarrow d = - 6 ∴ a_{20} = a + (20 - 1) d \Rightarrow a_{20} = a + 19 d \Rightarrow a_{20} = 2 + 19 (- 6) \Rightarrow a_{20} = - 112$

Page No 19.31:

Question 30:

If S₁ be the sum of (2n + 1) terms of an A.P. and S₂ be the sum of its odd terms, the prove that:
S₁ : S₂ = (2n + 1) : (n + 1)

Answer:

$Let the A . P . be a, a + d, a + 2 d . . . ∴ S_{1} = \frac{2 n + 1}{2} [2 a + (2 n + 1 - 1) d] \Rightarrow S_{1} = \frac{2 n + 1}{2} [2 a + (2 n) d] \Rightarrow S_{1} = (2 n + 1) (a + n d) . . . (i) S_{2} = \frac{n + 1}{2} [2 a + (n + 1 - 1) \times 2 d] \Rightarrow S_{2} = \frac{n + 1}{2} [2 a + 2 n d] \Rightarrow S_{2} = (n + 1) [a + n d] . . . (ii) From (i) and (ii), we get : \frac{S_{1}}{S_{2}} = \frac{2 n + 1}{n + 1} Hence, proved .$

Page No 19.31:

Question 31:

Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Answer:

Given:
$S_{n} = 3 n^{2} For n = 1, S_{1} = 3 \times 1^{2} = 3 For n = 2, S_{2} = 3 \times 2^{2} = 12 For n = 3, S_{3} = 3 \times 3^{2} = 27 and so on ∴ S_{1} = a_{1} = 3 a_{2} = S_{2} - S_{1} = 12 - 3 = 9 a_{3} = S_{3} - S_{2} = 27 - 12 = 15 and so on Thus, the A . P . is 3, 9, 15 . . .$

Page No 19.31:

Question 32:

If the sum of n terms of an A.P. is nP + $\frac{1}{2}$ n (n − 1) Q, where P and Q are constants, find the common difference.

Answer:

$W e have : S_{n} = n P + \frac{1}{2} n (n - 1) Q For n = 1, S_{1} = P + 0 = P For n = 2, S_{2} = 2 P + Q Also, a_{1} = S_{1} = P, a_{2} = S_{2} - S_{1} = 2 P + Q - P = P + Q ∴ d = a_{2} - a_{1} = P + Q - P = Q$

Page No 19.31:

Question 33:

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Answer:

$Let there be two A . P . s . Let their first terms be a_{1} and a_{2} and their common differences be d_{1} and d_{2} . Given : \frac{5 n + 4}{9 n + 6} = \frac{Sum of n terms in the first A . P .}{Sum of n terms in the second A . P .} \Rightarrow \frac{5 n + 4}{9 n + 6} = \frac{2 a_{1} + [(n - 1) d_{1}]}{2 a_{2} + [(n - 1) d_{2}]} Putting n = 2 \times 18 - 1 = 35 in the above equation, we get : \frac{5 \times 35 + 4}{9 \times 35 + 6} = \frac{2 a_{1} + 34 d_{1}}{2 a_{2} + 34 d_{2}} \Rightarrow \frac{179}{321} = \frac{a_{1} + 17 d_{1}}{a_{1} + 17 d_{1}} \Rightarrow \frac{179}{321} = \frac{18 th term of the first A . P .}{18 th term of the second A . P .}$

Page No 19.31:

Question 34:

The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.

Answer:

$Let the first term, the common difference and the sum of the first n terms of the first A . P . be a_{1}, d_{1} and S_{1}, respectively, and those of the second A . P . be a_{2}, d_{2} and S_{2}, respectively . Then, we have, S_{1} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}] And, S_{2} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}] Given : \frac{S_{1}}{S_{2}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 2}{n + 4} \Rightarrow \frac{S_{1}}{S_{2}} = \frac{[2 a_{1} + (n - 1) d_{1}]}{[2 a_{2} + (n - 1) d_{2}]} = \frac{7 n + 2}{n + 4} To find the ratio of the 5 th terms of the two A . P . s, we replace n by (2 \times 5 - 1 = 9) in the above equation : \Rightarrow \frac{[2 a_{1} + (9 - 1) d_{1}]}{[2 a_{2} + (9 - 1) d_{2}]} = \frac{7 \times 9 + 2}{9 + 4} \Rightarrow \frac{[2 a_{1} + (8) d_{1}]}{[2 a_{2} + (8) d_{2}]} = \frac{7 \times 9 + 2}{9 + 4} = \frac{65}{13} \Rightarrow \frac{[a_{1} + 4 d_{1}]}{[a_{2} + 4 d_{2}]} = \frac{5}{1} = 5 : 1$

Page No 19.4:

Question 1:

If the n^th term a_n of a sequence is given by a_n = n² − n + 1, write down its first five terms.

Answer:

$Given : a_{n} = n^{2} - n + 1 For n = 1, a_{1} = 1^{2} - 1 + 1 = 1 For n = 2, a_{2} = 2^{2} - 2 + 1 = 3 For n = 3, a_{3} = 3^{2} - 3 + 1 = 7 For n = 4, a_{4} = 4^{2} - 4 + 1 = 13 For n = 5, a_{5} = 5^{2} - 5 + 1 = 21$
Thus, the first five terms of the sequence are 1, 3, 7, 13, 21.

Page No 19.4:

Question 2:

A sequence is defined by a_n = n³ − 6n² + 11n − 6, n Ïµ N. Show that the first three terms of the sequence are zero and all other terms are positive.

Answer:

Given:
a_n = n³ − 6n² + 11n − 6, n Ïµ N
$For n = 1, a_{1} = 1^{3} - 6 \times 1^{2} + 11 \times 1 - 6 = 0 For n = 2, a_{2} = 2^{3} - 6 \times 2^{2} + 11 \times 2 - 6 = 0 For n = 3, a_{3} = 3^{3} - 6 \times 3^{2} + 11 \times 3 - 6 = 0 For n = 4, a_{4} = 4^{3} - 6 \times 4^{2} + 11 \times 4 - 6 = 6 > 0 For n = 5, a_{5} = 5^{3} - 6 \times 5^{2} + 11 \times 5 - 6 = 24 > 0 and so on Thus, the first three terms are zero and the rest of the terms are positive in the sequence .$

Page No 19.4:

Question 3:

Let < a_n > be a sequence defined by
a₁ = 3
and, a_n = 3a_n_{− 1} + 2, for all n > 1
Find the first four terms of the sequence.

Answer:

Given:
a₁ = 3
And, a_n = 3a_n_{− 1} + 2 for all n > 1

$a_{2} = 3 a_{2 - 1} + 2 = 3 a_{1} + 2 = 11 a_{3} = 3 a_{3 - 1} + 2 = 3 a_{2} + 2 = 35 a_{4} = 3 a_{4 - 1} + 2 = 3 a_{3} + 2 = 107 Thus, the first four terms of the sequence are 3, 11, 35, 107 .$

Page No 19.4:

Question 4:

Let < a_n > be a sequence. Write the first five terms in each of the following:
(i) a₁ = 1, a_n = a_n_{− 1} + 2, n ≥ 2
(ii) a₁ = 1 = a₂, a_n = a_n_{− 1} + a_n_{− 2}, n > 2
(iii) a₁ = a₂ = 2, a_n = a_n_{− 1} − 1, n > 2

Answer:

(i) a₁ = 1, a_n = a_n_{− 1} + 2, n ≥ 2
$a_{2} = a_{1} + 2 = 1 + 2 = 3 a_{3} = a_{2} + 2 = 5 a_{4} = a_{3} + 2 = 7 a_{5} = a_{4} + 2 = 9$

Hence, the five terms are 1, 3, 5, 7 and 9.

(ii) a₁ = 1 = a₂, a_n = a_n_{− 1} + a_n_{− 2}, n > 2
$a_{3} = a_{2} + a_{1} = 1 + 1 = 2 a_{4} = a_{3} + a_{2} = 2 + 1 = 3 a_{5} = a_{4} + a_{3} = 3 + 2 = 5$

Hence, the five terms are 1, 1, 2, 3 and 5.

(iii) a₁ = a₂ = 2, a_n = a_n_{− 1} − 1, n > 2
$a_{3} = a_{2} - 1 = 2 - 1 = 1 a_{4} = a_{3} - 1 = 1 - 1 = 0 a_{5} = a_{4} - 1 = 0 - 1 = - 1$

Hence, the five terms are 2, 2, 1, 0 and $-$ 1.

Page No 19.4:

Question 5:

The Fibonacci sequence is defined by
a₁ = 1 = a₂, a_n = a_n_{− 1} + a_n_{− 2} for n > 2

Find $\frac{{}^{a}n + 1}{{}^{a}n}$ for n = 1, 2, 3, 4, 5.

Answer:

a₁ = 1 = a₂, a_n = a_n_{− 1} + a_n_{− 2} for n > 2
Then, we have:
$a_{3} = a_{2} + a_{1} = 1 + 1 = 2 a_{4} = a_{3} + a_{2} = 2 + 1 = 3 a_{5} = a_{4} + a_{3} = 3 + 2 = 5 a_{6} = a_{5} + a_{4} = 5 + 3 = 8 For n = 1, \frac{a_{n + 1}}{a_{n}} = \frac{a_{2}}{a_{1}} = \frac{1}{1} = 1 For n = 2, \frac{a_{n + 1}}{a_{n}} = \frac{a_{3}}{a_{2}} = \frac{2}{1} = 2 For n = 3, \frac{a_{n + 1}}{a_{n}} = \frac{a_{4}}{a_{3}} = \frac{3}{2} For n = 4, \frac{a_{n + 1}}{a_{n}} = \frac{a_{5}}{a_{4}} = \frac{5}{3} For n = 5, \frac{a_{n + 1}}{a_{n}} = \frac{a_{6}}{a_{5}} = \frac{8}{5}$

Page No 19.4:

Question 6:

Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case.
(i) 3, −1, −5, −9 ...
(ii) −1, 1/4, 3/2, 11/4, ...
(iii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, . . .$
(iv) 9, 7, 5, 3, ...

Answer:

$(i) We have : - 1 - 3 = - 4, - 5 - (- 1) = - 4, - 9 - (- 5) = - 4 . . . Thus, the sequence is an A . P . with the common difference being - 4 . The next three terms are as follows : - 9 - 4 = - 13 - 13 - 4 = - 17 - 17 - 4 = - 21 (ii) We have : 1 / 4 - (- 1) = 5 / 4 3 / 2 - 1 / 4 = 5 / 4 11 / 4 - 3 / 2 = 5 / 4 Thus, the sequence is an A . P . with the common difference being (5 / 4) . The next three terms are as follows : 11 / 4 + 5 / 4 = 16 / 4 = 4 16 / 4 + 5 / 4 = 21 / 4 21 / 4 + 5 / 4 = 26 / 4$

$(iii) We have : 3 \sqrt{2} - \sqrt{2} = 2 \sqrt{2} 5 \sqrt{2} - 3 \sqrt{2} = 2 \sqrt{2} 7 \sqrt{2} - 5 \sqrt{2} = 2 \sqrt{2} Thus, the sequence is an A . P . with the common difference being (2 \sqrt{2}) . The next three terms are as follows : 7 \sqrt{2} + 2 \sqrt{2} = 9 \sqrt{2} 9 \sqrt{2} + 2 \sqrt{2} = 11 \sqrt{2} 11 \sqrt{2} + 2 \sqrt{2} = 13 \sqrt{2} (iv) We have : 7 - 9 = - 2 5 - 7 = - 2 3 - 5 = - 2 Thus, the sequence is an A . P . with the common difference being (- 2) . The next three terms are as follows : 3 - 2 = 1 1 - 2 = - 1 - 1 - 2 = - 3$

Page No 19.4:

Question 7:

The n^th term of a sequence is given by a_n = 2n + 7. Show that it is an A.P. Also, find its 7th term.

Answer:

$a_{n} = 2 n + 7 ∴ a_{1} = 2 \times 1 + 7 = 9 a_{2} = 2 \times 2 + 7 = 11 a_{3} = 2 \times 3 + 7 = 13 a_{4} = 2 \times 4 + 7 = 15 and so on So, common difference (d) = 11 - 9 = 2 Thus, the above sequence is an A . P . with the common difference as 2 a_{7} = 2 \times 7 + 7 = 21$

Page No 19.4:

Question 8:

The n^th term of a sequence is given by a_n = 2n² + n + 1. Show that it is not an A.P.

Answer:

We have:
a_n= 2n² + n + 1
$a_{1} = 2 \times 1^{2} + 1 + 1 = 4, a_{2} = 2 \times 2^{2} + 2 + 1 = 11 a_{3} = 2 \times 3^{2} + 3 + 1 = 22 a_{2} - a_{1} = 11 - 4 = 7 and a_{3} - a_{2} = 22 - 11 = 11 Since, a_{2} - a_{1} \neq a_{3} - a_{2} Hence, it is not an AP .$

Page No 19.42:

Question 1:

If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., prove that:
(i) $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P.
(ii) a (b +c), b (c + a), c (a +b) are in A.P.

Answer:

$Given : \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in A . P . ∴ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \Rightarrow 2 a c = a b + b c . . . . (1) (i) To prove : \frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c} are in A . P . 2 (\frac{a + c}{b}) = \frac{b + c}{a} + \frac{a + b}{c} \Rightarrow 2 a c (a + c) = b c (b + c) + a b (a + b) LHS : 2 a c (a + c) = (a b + b c) (a + c) (From (1)) = a^{2} b + 2 a b c + b c^{2} RHS : b c (b + c) + a b (a + b) = b^{2} c + b c^{2} + a^{2} b + a b^{2} = b^{2} c + a b^{2} + b c^{2} + a^{2} b = b (b c + a b) + b c^{2} + a^{2} b = 2 a b c + b c^{2} + a^{2} b = a^{2} b + 2 a b c + b c^{2} (From (1)) ∴ LHS = RHS Hence, proved .$

$(ii) To prove : a (b + c), b (c + a), c (a + b) are in A . P . \Rightarrow 2 b (c + a) = a (b + c) + c (a + b) LHS : 2 b (c + a) = 2 b c + 2 b a RHS : a (b + c) + c (a + b) = a b + a c + a c + b c = a b + 2 a c + b c = a b + a b + b c + b c (From (1)) = 2 a b + 2 b c ∴ LHS = RHS Hence, proved .$

Page No 19.42:

Question 2:

If a², b², c² are in A.P., prove that $\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}$ are in A.P.

Answer:

$a^{2}, b^{2}, c^{2} are in A . P . ∴ 2 b^{2} = a^{2} + c^{2} \Rightarrow b^{2} - a^{2} = c^{2} - b^{2} \Rightarrow (b + a) (b - a) = (c - b) (c + b) \Rightarrow \frac{b - a}{c + b} = \frac{c - b}{b + a} \Rightarrow \frac{b - a}{(c + a) (c + b)} = \frac{c - b}{(b + a) (c + a)} [Multiplying both the sides by \frac{1}{c + a}] \Rightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a} ∴' \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b} are in A . P . Multiplying each term by (a + b + c) : \frac{a + b + c}{b + c}, \frac{a + b + c}{c + a}, \frac{a + b + c}{a + b} are in A . P . Thus, \frac{a}{b + c} + 1, \frac{b}{c + a} + 1, \frac{c}{a + b} + 1 are in A . P . Hence, \frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b} are in A . P .$

Page No 19.42:

Question 3:

If a, b, c are in A.P., then show that:
(i) a² (b + c), b² (c + a), c² (a + b) are also in A.P.
(ii) b + c − a, c + a − b, a + b − c are in A.P.
(iii) bc − a², ca − b², ab − c² are in A.P.

Answer:

$Since a, b, c are in A . P ., we have : 2 b = a + c (i) We have to prove the following : 2 b^{2} (a + c) = a^{2} (b + c) + c^{2} (a + b) LHS : 2 b^{2} \times 2 b (Given) = 4 b^{3} RHS : a^{2} b + a^{2} c + a c^{2} + c^{2} b = a c (a + c) + b (a^{2} + c^{2}) = a c (a + c) + b [(a + c)^{2} - 2 a c] = a c (2 b) + b [{(2 b)}^{2} - 2 a c] = 2 a b c + 4 b^{3} - 2 a b c = 4 b^{3} LHS = RHS Hence, proved .$

$(ii) We have to prove the following : 2 (c + a - b) = (b + c - a) + (a + b - c) LHS : 2 (c + a - b) = 2 (2 b - b) (∵ 2 b = a + c) = 2 b RHS : (b + c - a) + (a + b - c) = 2 b LHS = RHS Hence, proved .$

$(iii) We have to prove t h e f o l l o w i n g : 2 (c a - b^{2}) = (b c - a^{2} + a b - c^{2}) RHS : b c - a^{2} + a b - c^{2} = c (b - c) + a (b - a) = c (\frac{a + c}{2} - c) + a (\frac{a + c}{2} - a) (∵ 2 b = a + c) = c (\frac{a + c - 2 c}{2}) + a (\frac{a + c - 2 a}{2}) = \frac{c (a - c)}{2} + a (\frac{c - a}{2}) = \frac{c a}{2} - \frac{c^{2}}{2} + \frac{a c}{2} - \frac{a^{2}}{2} = a c - \frac{1}{2} (c^{2} + a^{2}) = a c - \frac{1}{2} (4 b^{2} - 2 a c) (∵ a^{2} + c^{2} + 2 a c = 4 b^{2} \Rightarrow a^{2} + c^{2} = 4 b^{2} - 2 a c) = a c - 2 b^{2} + a c = 2 a c - 2 b^{2} = 2 (a c - b^{2}) = LHS Hence, proved .$

Page No 19.42:

Question 4:

If $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P., prove that:

(i) $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

(ii) bc, ca, ab are in A.P.

Answer:

(i) Since $\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}$ are in A.P., we have:

$\frac{c + a}{b} - \frac{b + c}{a} = \frac{a + b}{c} - \frac{c + a}{b} \Rightarrow \frac{a c + a^{2} - b^{2} - b c}{a b} = \frac{a b + b^{2} - c^{2} - a c}{b c} \Rightarrow \frac{(a + b) (a - b) + c (a - b)}{a b} = \frac{(b + c) (b - c) + a (b - c)}{b c} \Rightarrow \frac{(a - b) (a + b + c)}{a b} = \frac{(b - c) (a + b + c)}{b c} \Rightarrow \frac{(a - b)}{a b} = \frac{(b - c)}{b c} \Rightarrow \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}$

Hence, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

(ii) Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., we have:

$\frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \Rightarrow \frac{(a - b)}{a b} = \frac{(b - c)}{b c} \Rightarrow \frac{(a - b)}{a} = \frac{(b - c)}{c} \Rightarrow (a - b) c = a (b - c) \Rightarrow a c - b c = a b - a c$

Hence, bc, ca, ab are in A.P.

Page No 19.42:

Question 5:

If a, b, c are in A.P., prove that:
(i) (a − c)² = 4 (a − b) (b − c)
(ii) a² + c² + 4ac = 2 (ab + bc + ca)
(iii) a³ + c³ + 6abc = 8b³.

Answer:

Since a, b, c are in A.P., we have:
2b = a+c
$\Rightarrow$ b = $\frac{a + c}{2}$

(i) Consider RHS:
4 (a − b) (b − c)
$Substituting b = \frac{a + c}{2} : \Rightarrow 4 \{a - \frac{a + c}{2}\} \{\frac{a + c}{2} - c\} \Rightarrow 4 \{\frac{2 a - a - c}{2}\} \{\frac{a + c - 2 c}{2}\} \Rightarrow (a - c) (a - c) \Rightarrow {(a - c)}^{2}$
Hence, proved.

(ii) Consider RHS:
2 (ab + bc + ca)
$Substituting b = \frac{a + c}{2} : \Rightarrow 2 \{a (\frac{a + c}{2}) + c (\frac{a + c}{2}) + a c\} \Rightarrow 2 \{\frac{a^{2} + a c + a c + c^{2} + 2 a c}{2}\} \Rightarrow a^{2} + 4 a c + c^{2} Hence, proved .$

(iii) Consider RHS:
8 $b^{3}$
$Substituting b = \frac{a + c}{2} : \Rightarrow 8 {(\frac{a + c}{2})}^{3} \Rightarrow a^{3} + c^{3} + 3 a c (a + c) \Rightarrow a^{3} + c^{3} + 3 a c (2 b) \Rightarrow a^{3} + c^{3} + 6 a b c$

Hence, proved.

Page No 19.42:

Question 6:

If $a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}), c (\frac{1}{a} + \frac{1}{b})$ are in A.P., prove that a, b, c are in A.P.

Answer:

Given:
$a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}), c (\frac{1}{a} + \frac{1}{b})$ are in A.P.
$By adding 1 to each term, we get : a (\frac{1}{b} + \frac{1}{c}) + 1, b (\frac{1}{c} + \frac{1}{a}) + 1, c (\frac{1}{a} + \frac{1}{b}) + 1 are in A . P . \Rightarrow a (\frac{1}{b} + \frac{1}{c} + \frac{1}{a}), b (\frac{1}{c} + \frac{1}{a} + \frac{1}{b}), c (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) are in A . P . Dividing all terms by \frac{1}{a} + \frac{1}{b} + \frac{1}{c}, we get : \Rightarrow a, b, c are in A . P . Hence, proved .$

Page No 19.42:

Question 7:

Show that x² + xy + y², z² + zx + x² and y² + yz + z² are consecutive terms of an A.P., if x, y and z are in A.P. [NCERT EXEMPLAR]

Answer:

$As, x, y and z are in A . P . So, y = \frac{x + z}{2} . . . . . (i) Now, (x^{2} + x y + y^{2}) + (y^{2} + y z + z^{2}) = x^{2} + z^{2} + 2 y^{2} + x y + y z = x^{2} + z^{2} + 2 y^{2} + y (x + z) = x^{2} + z^{2} + 2 {(\frac{x + z}{2})}^{2} + (\frac{x + z}{2}) (x + z) [Using (i)] = x^{2} + z^{2} + 2 (\frac{{(x + z)}^{2}}{4}) + \frac{{(x + z)}^{2}}{2} = x^{2} + z^{2} + \frac{{(x + z)}^{2}}{2} + \frac{{(x + z)}^{2}}{2} = x^{2} + z^{2} + {(x + z)}^{2} = x^{2} + z^{2} + x^{2} + 2 x y + z^{2} = 2 x^{2} + 2 x y + 2 z^{2} = 2 (x^{2} + x y + z^{2}) Since, (x^{2} + x y + y^{2}) + (y^{2} + y z + z^{2}) = 2 (x^{2} + x y + z^{2}) So, (x^{2} + x y + y^{2}), (x^{2} + x y + z^{2}) and (y^{2} + y z + z^{2}) are in A . P .$

Hence, x² + xy + y², z² + zx + x² and y² + yz + z² are consecutive terms of an A.P.

Page No 19.46:

Question 1:

Find the A.M. between:
(i) 7 and 13
(ii) 12 and −8
(iii) (x − y) and (x + y).

Answer:

(i) Let $A_{1}$ be the A.M. between 7 and 13.

$A_{1}$ = $\frac{a + b}{2}$

     = $\frac{7 + 13}{2}$

     = 10

(ii) Let $A_{1}$ be the A.M. between 12 and −8.

$A_{1}$ = $\frac{a + b}{2}$
    = $\frac{12 + (- 8)}{2}$
    = 2

(iii) Let $A_{1}$ be the A.M. between (x − y) and (x + y).
$A_{1}$ = $\frac{a + b}{2}$
    = $\frac{(x - y) + (x + y)}{2}$
    = x

Page No 19.46:

Question 2:

Insert 4 A.M.s between 4 and 19.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the four A.M.s between 4 and 19. Then, 4, $A_{1}, A_{2}, A_{3}, A_{4}$ and 19 are in A.P. whose common difference is as follows:
d = $\frac{19 - 4}{4 + 1}$
= 3

$A_{1} = 4 + d = 4 + 3 = 7 A_{2} = 4 + 2 d = 4 + 6 = 10 A_{3} = 4 + 3 d = 4 + 9 = 13 A_{4} = 4 + 4 d = 4 + 12 = 16$
Hence, the required A.M.s are 7, 10, 13, 16.

Page No 19.46:

Question 3:

Insert 7 A.M.s between 2 and 17.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ be the seven A.M.s between 2 and 17.
Then, 2, $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ and 17 are in A.P. whose common difference is as follows:

d = $\frac{17 - 2}{7 + 1}$

= $\frac{15}{8}$

$A_{1} = 2 + d = 2 + \frac{15}{8} = \frac{31}{8} A_{2} = 2 + 2 d = 2 + \frac{15}{4} = \frac{23}{4} A_{3} = 2 + 3 d = 2 + \frac{45}{8} = \frac{61}{8} A_{4} = 2 + 4 d = 2 + \frac{15}{2} = \frac{19}{2} A_{5} = 2 + 5 d = 2 + \frac{75}{8} = \frac{91}{8} A_{6} = 2 + 6 d = 2 + \frac{45}{4} = \frac{53}{4} A_{7} = 2 + 7 d = 2 + \frac{105}{8} = \frac{121}{8}$

Hence, the required A.M.s are $\frac{31}{8}, \frac{23}{4}, \frac{61}{8}, \frac{19}{2}, \frac{91}{8}, \frac{53}{4}, \frac{121}{8}$ .

Page No 19.46:

Question 4:

Insert six A.M.s between 15 and −13.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be the 6 A.M.s between 15 and $-$ 13.
Then, 15, $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ and $-$ 13 are in A.P. whose common difference is as follows:

d = $\frac{- 13 - 15}{6 + 1}$
= $- 4$

$A_{1} = 15 + d = 15 + (- 4) = 11 A_{2} = 15 + 2 d = 15 + (- 8) = 7 A_{3} = 15 + 3 d = 15 + (- 12) = 3 A_{4} = 15 + 4 d = 15 + (- 16) = - 1 A_{5} = 15 + 5 d = 15 + (- 20) = - 5 A_{6} = 15 + 6 d = 15 + (- 24) = - 9$
Hence, the required A.M.s are $11, 7, 3, - 1, - 5, - 9$ .

Page No 19.46:

Question 5:

There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value of n.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4} . . . . A_{n}$ be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, $A_{1}, A_{2}, A_{3}, A_{4} . . . . A_{n}$ and17.
Then, we have:
d = $\frac{17 - 3}{n + 1}$ = $\frac{14}{n + 1}$

Now, $A_{1}$ = 3 + d = 3 + $\frac{14}{n + 1}$ = $\frac{3 n + 17}{n + 1}$

And, $A_{n} = 3 + n d = 3 + n (\frac{14}{n + 1}) = \frac{17 n + 3}{n + 1}$

$∴ \frac{A_{n}}{A_{1}} = \frac{3}{1} \Rightarrow \frac{(\frac{17 n + 3}{n + 1})}{(\frac{3 n + 17}{n + 1})} = \frac{3}{1} \Rightarrow \frac{17 n + 3}{3 n + 17} = \frac{3}{1} \Rightarrow 17 n + 3 = 9 n + 51 \Rightarrow 8 n = 48 \Rightarrow n = 6$

Page No 19.46:

Question 6:

Insert A.M.s between 7 and 71 in such a way that the 5^th A.M. is 27. Find the number of A.M.s.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4}, 27, A_{6} . . . A_{n}$ be the n arithmetic means between 7 and 71.
Thus, there are $(n + 2)$ terms in all.
Let d be the common difference of the above A.P.
Now, a₆ = 27
$\Rightarrow a + (6 - 1) d = 27 \Rightarrow a + 5 d = 27$
$\Rightarrow$ d = 4

Also, 71 = a_n+2
71 = 7+ $(n + 2 - 1)$ 4
$\Rightarrow$ 71 = 7+ $(n + 1)$ 4
$\Rightarrow$ n = 15

Therefore, there are 15 A.M.s.

Page No 19.46:

Question 7:

If n A.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Answer:

$Let A_{1}, A_{2} . . . . . . A_{n} be n A . M . s between two numbers a and b . Then, a, A_{1}, A_{2} . . . . . . . A_{n}, b are in A . P . with common difference, d = \frac{b - a}{n + 1} . ∴ A_{1} + A_{2} + . . . . . . + A_{n} = \frac{n}{2} [A_{1} + A_{n}] = \frac{n}{2} [A_{1} - d + A_{n} + d] = \frac{n}{2} [a + b] = n \times [\frac{a + b}{2}] = A . M . between a and b, which is constant .$

Page No 19.47:

Question 8:

If x, y, z are in A.P. and A₁ is the A.M. of x and y and A₂ is the A.M. of y and z, then prove that the A.M. of A₁ and A₂ is y.

Answer:

x, y, z are in A.P.
$∴$ $y = \frac{x + z}{2}$

Now, $A_{1} is the arithmetic mean of x and y .$
$A_{1} = \frac{x + y}{2} = \frac{x + \frac{x + z}{2}}{2} = \frac{3 x + z}{4}$

And, $A_{2} is the arithmetic mean of y and z .$
$A_{2} = \frac{y + z}{2} = \frac{\frac{x + z}{2} + z}{2} = \frac{3 z + x}{4}$

Let $A_{3}$ be the arithmetic mean of $A_{1} and A_{2}$ .

$A_{3} = \frac{A_{1} + A_{2}}{2} = \frac{\frac{3 x + z}{4} + \frac{3 z + x}{4}}{2} = \frac{4 x + 4 z}{8} = \frac{x + z}{2} = y$

Hence, proved.

Page No 19.47:

Question 9:

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer:

Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ be five numbers between 8 and 26.
Let d be the common difference.
Then, we have:
26 = a₇
$\Rightarrow$ 26 = 8 + $(7 - 1)$ d
$\Rightarrow$ 26 = 8 + 6d
$\Rightarrow$ d = 3

$A_{1} = 8 + d = 8 + 3 = 11 A_{2} = 8 + 2 d = 8 + 6 = 14 A_{3} = 8 + 3 d = 8 + 9 = 17 A_{4} = 8 + 4 d = 8 + 12 = 20 A_{5} = 8 + 5 d = 8 + 15 = 23$

Therefore, the five numbers are 11, 14, 17, 20, 23.

Page No 19.48:

Question 6:

A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

Answer:

It is given that the man counts Rs 180 per minute for half an hour.
∴ Sum of money the man counts in 30 minutes = Rs 180 $\times$ 30 = Rs 5400

Total money counted by the man = Rs 10710
∴ Money left for counting after 30 minutes = Rs (10710 − 5400) = Rs 5310

It is given that after 30 minutes, he counts at the rate of Rs 3 less every minute than the preceding minute.

Therefore, it would be an A.P. where a = 177 and d = −3.
Let the time taken to count Rs 5310 be n minutes.
$5310 = \frac{n}{2} [2 \times 177 + (n - 1) \times - 3] \Rightarrow 10620 = 354 n - 3 n^{2} + 3 n \Rightarrow 3 n^{2} - 357 n + 10620 = 0 \Rightarrow n^{2} - 119 n + 3540 = 0 \Rightarrow n^{2} - 59 n - 60 n + 3540 = 0 \Rightarrow n (n - 59) - 60 (n - 59) = 0 \Rightarrow (n - 59) (n - 60) = 0 ∴ n = 59 or 60$

Thus, the time taken to count Rs 5310 would be 59 minutes or 60 minutes.

Hence, the total time taken to count Rs 10710 would be (30 + 59) minutes or (30 + 60) minutes, i.e. 89 minutes or 90 minutes, respectively.

Page No 19.49:

Question 1:

A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the receding year. How much did he save in the first year?

Answer:

Let the amount saved by the man in the first year be Rs A.
Let d be the common difference.
Let $S_{10}$ denote the amount he saves in ten years.
Here, n =10, d =100

We know:
$S_{n} = \frac{n}{2} \{2 A + (n - 1) d\} ∴ S_{10} = \frac{10}{2} \{2 A + (10 - 1) 100\} \Rightarrow 16500 = 5 \{2 A + 900\} \Rightarrow 3300 = 2 A + 900 \Rightarrow A = 1200$

Therefore, the man saved Rs 1200 in the first year.

Page No 19.49:

Question 2:

A man saves Rs 32 during the first year. Rs 36 in the second year and in this way he increases his savings by Rs 4 every year. Find in what time his saving will be Rs 200.

Answer:

Let n be the time in which the man saved Rs 200.
Here, d= 4, a = 32
We know:
$S_{n} = \frac{n}{2} \{2 a + (n - 1) d\} \Rightarrow 200 = \frac{n}{2} \{2 \times 32 + (n - 1) 4\} \Rightarrow 400 = 64 n + 4 n^{2} - 4 n \Rightarrow 4 n^{2} + 60 n - 400 = 0 \Rightarrow n^{2} + 15 n - 100 = 0 \Rightarrow n^{2} + 20 n - 5 n - 100 = 0 \Rightarrow (n + 20) (n - 5) = 0 \Rightarrow n = 5, n = - 20 (Rejecting the negative value)$

Therefore, the man took 5 years to save Rs 200.

Page No 19.49:

Question 3:

A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.

Answer:

Let $S_{40}$ denote the total loan amount to be paid in 40 annual instalments.
∴ $S_{40}$ = 3600
Let Rs a be the value of the first instalment and Rs d be the common difference.
We know:

$S_{n} = \frac{n}{2} \{2 a + (n - 1) d\}$

$\Rightarrow \frac{40}{2} \{2 a + (40 - 1) d\} = 3600 \Rightarrow 20 \{2 a + 39 d\} = 3600 \Rightarrow 2 a + 39 d = 180 . . . . . . (1)$

$Also, S_{40} - S_{30} = \frac{1}{3} \times 3600 \Rightarrow 3600 - S_{30} = 1200 \Rightarrow S_{30} = 2400 \Rightarrow \frac{30}{2} \{2 a + (30 - 1) d\} = 2400 \Rightarrow 15 \{2 a + 29 d\} = 2400 \Rightarrow 2 a + 29 d = 160 . . . . . (2)$

On solving equations $(1) and (2)$ , we get:
d = 2 and a =51

Hence, the value of the first instalment is Rs 51

Page No 19.49:

Question 4:

A manufacturer of radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find (i) the production in the first year (ii) the total product in 7 years and (iii) the product in the 10th year.

Answer:

Let $a_{n}$ denote the production of radio sets in the nth year.
Here, $a_{3}$ = 600, $a_{7}$ = 700
We know:
$a_{n} = a + (n - 1) d$

$a_{3} = a + 2 d \Rightarrow 600 = a + 2 d . . . . . (1)$

$And, a_{7} = a + 6 d \Rightarrow 700 = a + 6 d . . . . . (2)$

Solving $(1)$ and $(2)$ , we get:
d = 25, a = 550
Hence, the production in the first year is 550 units.

(ii) Let $S_{n}$ denote the total production in n years.
Total production in 7 years = $S_{7}$
$= \frac{7}{2} \{2 \times 550 + (7 - 1) 25\} = 4375 units$

(iii) Production in the 10th year = $a_{10}$
$a_{10} = a + (10 - 1) d = 550 + 9 (25) = 775$

Page No 19.49:

Question 5:

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Answer:

Let $S_{n}$ be the total distance travelled by the gardener.
Let d be the common difference (distance) between two trees. Let a be the distance of the well from the first tree.
Here, n = 25, d = 10, a = 20
Distance travelled by the gardener from the well to the last tree = $S_{25}$
$S_{25} = \frac{25}{2} \{2 \times 20 + (25 - 1) 10\} = \frac{25}{2} (40 + 240) = 3500 m$

Therefore, the total distance the gardener has to travel is 3500 m.

Page No 19.49:

Question 7:

A piece of equipment cost a certain factory Rs 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Answer:

Cost of the piece of equipment at the end of the first year = Rs 600000 – 15% of 600000
                                                                                      = Rs 600000 – Rs 90000
                                                            = Rs 510000
Cost of the piece of equipment at the end of the second year = Rs 600000 – 13.5% of 600000
                                                                    = Rs 600000 – Rs 81000
                                                                  = Rs 519000
Cost of the piece of equipment at the end of the third year = Rs 600000 – 12% of Rs 600000
                                                                  = Rs 600000 – Rs 72000
                                                                = Rs 528000
The numbers 510000, 519000, 528000 are in an A.P. whose first term in 510000 and common difference is 9000.
∴ Cost of the piece of equipment at the end of 10 years = a + (10 – 1)d
                                                                    = 510000 + 81000
                                                                = Rs 591000

Page No 19.49:

Question 8:

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. How much the tractor cost him?

Answer:

Cost of the tractor = Rs 12000
It is given that the farmer pays Rs 6000 in cash.
Unpaid amount = Rs 6000
He has to pay Rs 6000 in annual instalments of Rs 500 plus 12% interest on the unpaid amount.
∴ Number of years taken by the farmer to pay the whole amount = 6000 $\div$ 500 = 12
Hence, the interest paid by farmer annually would be as follows:
$12 % of Rs 6000 + 12 % of Rs 5500 + 12 % of Rs 5000 . . = 720 + 660 + 600 . . . .$

It is in an A.P. where a = 720 , d = $-$ 60 and n = 12.
Total sum:
$\frac{12}{2} [2 \times 720 + 11 \times - 60] = 6 [1440 - 660] = Rs 4680$

∴ Amount the farmer has to pay = Rs 12000 + Rs 4680 = Rs 16680

Page No 19.49:

Question 9:

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount. How much the scooter will cost him.

Answer:

Cost of the scooter = Rs 22000
Shamshad Ali pays Rs 4000 in cash.
∴ Unpaid amount = Rs 22000 $-$ Rs 4000 = Rs 18000
Number of years taken by Shamshed Ali to pay the whole amount = 18000 $\div$ 1000 = 18
He agrees to pay the balance in annual instalments of Rs 1000 plus 10% interest on the unpaid amount.

Total amount of instalments:
$10 % of Rs 18000 + 10 % of Rs 17000 + 10 % of Rs 16000 . . . . = 1800 + 1700 + 1600 . . . .$

It is in an A.P. where a = 1800, d = $-$ 100 and n = 18.
Therefore, total amount of instalments:
$\frac{18}{2} [2 \times 1800 + (18 - 1) \times - 100] = 9 [3600 - 1700] = Rs 17100$

∴ Total amount Shamshad Ali has to pay = Rs (22000 + 17100) = Rs 39100

Page No 19.49:

Question 10:

The income of a person is Rs 300,000 in the first year and he receives an increase of Rs 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.

Answer:

Let $S_{n}$ denote the total amount the person receives in n years.
Let d be the common increment in his income every year.
Let a denote the initial income of the person.
Here, a = 300,000, d = 10000, n = 20

Total amount at the end of 20 years:
$S_{20} = \frac{20}{2} \{2 \times 300, 000 + (20 - 1) 10, 000\} = 79, 00, 000$

Therefore, the total amount the person receives in 20 years is Rs 79,00,000.

Page No 19.49:

Question 11:

A man starts repaying a loan as first instalment of Rs 100 = 00. If he increases the instalments by Rs 5 every month, what amount he will pay in the 30th instalment?

Answer:

Let $a_{30}$ be the amount a man repays in the 30th instalment.
Let d be the common increment in his instalment every month.
Let a be the initial repayment.
Here, a = 100, d = 5, n = 30
Amount to be repaid in the 30th instalment:
$a_{30}$ $\Rightarrow$ a+ $(n - 1)$ d
$= 100 + 29 \times 5 = 245$

Hence, the man repays Rs 245 in his 30th instalment.

Page No 19.49:

Question 12:

A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job? [NCERT EXEMPLAR]

Answer:

$We have, S = 192, a = 5, d = 2 Now, S_{n} = 192 \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = 192 \Rightarrow \frac{n}{2} [2 \times 5 + (n - 1) \times 2] = 192 \Rightarrow \frac{n}{2} [10 + 2 n - 2] = 192 \Rightarrow \frac{n}{2} [2 n + 8] = 192 \Rightarrow n (n + 4) = 192 \Rightarrow n^{2} + 4 n = 192 \Rightarrow n^{2} - 12 n + 16 n - 192 = 0 \Rightarrow n (n - 12) + 16 (n - 12) = 0 \Rightarrow (n - 12) (n + 16) = 0 \Rightarrow (n - 12) = 0 or (n + 16) = 0 \Rightarrow n = 12 or n = - 16 ∵ n cannot be negative . ∴ n = 12$

So, the carpenter takes 12 days to finish the job.

Page No 19.50:

Question 13:

We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. [NCERT EXEMPLAR]

Answer:

We know that,
the sum of the interior angles of a polygon with 3 sides, a₁ = 180°,
the sum of the interior angles of a polygon with 4 sides, a₂ = 360°,
the sum of the interior angles of a polygon with 5 sides, a₃ = 540°,
.
.
.

$As, a_{2} - a_{1} = 360 ° - 180 ° = 180 ° and a_{3} - a_{2} = 540 ° - 360 ° = 180 ° i . e . a_{2} - a_{1} = a_{3} - a_{2} So, a_{1}, a_{2}, a_{3}, . . . are in A . P . Also, a = 180 ° and d = 180 ° Since, the sum of the interior angles of a 3 sided polygon = a So, the sum of the interior angles of a 21 sided polygon = a_{19} Now, a_{19} = a + (19 - 1) d = 180 ° + 18 \times 180 ° = 180 ° + 3240 ° = 3420 °$

So, the sum of the interior angles for a 21 sided polygon is 3420°.

Page No 19.50:

Question 14:

In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
[NCERT EXEMPLAR]

Answer:

We have,

the distance travelled to bring the first potato, a₁ = 2 $\times$ 24 = 48 m,
the distance travelled to bring the second potato, a₂ = 2 $\times$ (24 + 4) = 56 m,
the distance travelled to bring the third potato, a₃ = 2 $\times$ (24 + 4 + 4) = 64 m,
.
.
.

$As, a_{2} - a_{1} = 56 - 48 = 8 and a_{3} - a_{2} = 64 - 56 = 8 i . e . a_{2} - a_{1} = a_{3} - a_{2} So, a_{1}, a_{2}, a_{3}, . . . are in A . P . Also, a = 48, d = 8, n = 20 Now, S_{20} = \frac{20}{2} [2 a + (20 - 1) d] = 10 [2 \times 48 + 19 \times 8] = 10 \times (96 + 152) = 10 \times 248 = 2480$

So, he would have to run 2480 m to bring back all the potatoes.

Page No 19.50:

Question 15:

A man accepts a position with an initial salary of â‚¹5200 per month. It is understood that he will receive an automatic increase of â‚¹320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?

Answer:

We have,

the initial salary, a₁ = â‚¹5200,
the salary of the second month, a₂ = â‚¹5200 + â‚¹320 = â‚¹5520,
the salary of the third month, a₃ = â‚¹5520 + â‚¹320 = â‚¹5840,
.
.
.

$As, a_{2} - a_{1} = 5520 - 5200 = 320 a n d a_{3} - a_{2} = 5840 - 5520 = 320 i . e . a_{2} - a_{1} = a_{3} - a_{2} So, a_{1}, a_{2}, a_{3}, . . . are in A . P . Also, a = 5200, d = 320 (i) a_{10} = a + (10 - 1) d = 5200 + 9 \times 320 = 5200 + 2880 = 8080 So, the salary of the man for the tenth month is ₹ 8, 080 . (ii) S_{12} = \frac{12}{2} [2 a + (12 - 1) d] = 6 (2 \times 5200 + 11 \times 320) = 6 (10400 + 3520) = 6 \times 13920 = 83520 So, the total earnings of the man during the first year is ₹ 83, 520 .$

Page No 19.50:

Question 16:

A man saved â‚¹66000 in 20 years. In each succeeding year after the first year he saved â‚¹200 more than what he saved in the previous year. How much did he save in the first year?

Answer:

As, in each succeeding year after the first year he saved â‚¹200 more than what he saved in the previous year.
So, the savings of each year are in A.P.

We have,
the total savings of the man in 20 years, S₂₀ = â‚¹66000 and
the difference of his savings in each succeeding year, d = â‚¹200

Let his savings in the first year be a.

Now,
$S_{20} = 66000 \Rightarrow \frac{20}{2} [2 a + (20 - 1) d] = 66000 \Rightarrow 10 [2 a + 19 \times 200] = 66000 \Rightarrow 2 a + 3800 = \frac{66000}{10} \Rightarrow 2 a = 6600 - 3800 \Rightarrow a = \frac{2800}{2} ∴ a = 1400$

So, he saved â‚¹1400 in the first year.

Page No 19.50:

Question 17:

In a cricket team tournament 16 teams participated. A sum of â‚¹8000 is to be awarded among themselves as prize money. If the last place team is awarded â‚¹275 in prize money and the award increases by the same amount for successive finishing places, then how much amount will the first place team receive?

Answer:

We have,
the total sum of prize money to be awarded among 16 teams, S₁₆ = â‚¹8000 and
the prize money awarded to the last place team i.e. a₁₆ = â‚¹275

As, the award increases by the same amount for successive finishing places.
So, the prize money are in A.P.

Let the prize money awarded to the first team be a.

Now,

$S_{16} = 8000 \Rightarrow \frac{16}{2} [a + a_{16}] = 8000 \Rightarrow 8 [a + 275] = 8000 \Rightarrow a + 275 = \frac{8000}{8} \Rightarrow a = 1000 - 275 ∴ a = 725$

So, the amount which the first place team will recieve is â‚¹725.

Page No 19.50:

Question 1:

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90

Answer:

(c) 89

$a_{7} = 34 \Rightarrow a + 6 d = 34 . . . . . (1) Also, a_{13} = 64 \Rightarrow a + 12 d = 64 . . . . . (2)$

Solving equations $(1) and (2)$ , we get:

a = 4 and d = 5

$∴ a_{18} = a + 17 d = 4 + 17 (5) = 89$

Page No 19.50:

Question 2:

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be
(a) 0
(b) p − q
(c) p + q
(d) − (p + q)

Answer:

(d) $- (p + q)$

$S_{p} = q \Rightarrow \frac{p}{2} \{2 a + (p - 1) d\} = q \Rightarrow 2 a p + (p - 1) p d = 2 q . . . . . (1) S_{q} = p \Rightarrow \frac{q}{2} \{2 a + (q - 1) d\} = p \Rightarrow 2 a q + (q - 1) q d = 2 p . . . . . (2) Multiplying equation (1) by q and equation (2) by p and then solving, we get : d = \frac{- 2 (p + q)}{p q} Now, S_{p + q} = \frac{(p + q)}{2} [2 a + (p + q - 1) d] = \frac{p}{2} [2 a + (p - 1) d + q d] + \frac{q}{2} [2 a + (q - 1) d + p d] = S_{p} + \frac{p q d}{2} + S_{q} + \frac{p q d}{2} = p + q + p q d = p + q - \frac{2 (p + q) p q}{p q} = - (p + q)$

Page No 19.50:

Question 3:

If the sum of n terms of an A.P. be 3 n² − n and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4

Answer:

(a) 2

$S_{n} = 3 n^{2} - n \Rightarrow S_{1} = 3 {(1)}^{2} - 1 \Rightarrow S_{1} = 2 ∴ a_{1} = 2$

Page No 19.50:

Question 4:

Sum of all two digit numbers which when divided by 4 yield unity as remainder is
(a) 1200
(b) 1210
(c) 1250
(d) none of these.

Answer:

(b) 1210

The given series is 13, 17, 21....97.

$a_{1} = 13, a_{2} = 17, a_{n} = 97 d = a_{2} - a_{1} = 7 - 3 = 4$

$a_{n} = 97 \Rightarrow a + (n - 1) d = 97 \Rightarrow 13 + (n - 1) 4 = 97 \Rightarrow n = 22$

Sum of the above series:
$S_{22} = \frac{22}{2} \{2 \times 13 + (22 - 1) 4\} = 11 \{26 + 84\} = 1210$

Page No 19.51:

Question 5:

In n A.M.'s are introduced between 3 and 17 such that the ratio of the last mean to the first mean is 3 : 1, then the value of n is
(a) 6
(b) 8
(c) 4
(d) none of these.

Answer:

(a) 6
Let $A_{1}, A_{2}, A_{3}, A_{4} . . . . A_{n}$ be the n arithmetic means between 3 and 17.
Let d be the common difference of the A.P. 3, $A_{1}, A_{2}, A_{3}, A_{4}, . . . . A_{n}$ and 17.
Then, we have:

d = $\frac{17 - 3}{n + 1}$ = $\frac{14}{n + 1}$

Now, $A_{1}$ = 3 + d = 3 + $\frac{14}{n + 1}$ = $\frac{3 n + 17}{n + 1}$

And, $A_{n} = 3 + n d = 3 + n (\frac{14}{n + 1}) = \frac{17 n + 3}{n + 1}$

$∴ \frac{A_{n}}{A_{1}} = \frac{3}{1} \Rightarrow \frac{(\frac{17 n + 3}{n + 1})}{(\frac{3 n + 17}{n + 1})} = \frac{3}{1} \Rightarrow \frac{17 n + 3}{3 n + 17} = \frac{3}{1} \Rightarrow 17 n + 3 = 9 n + 51 \Rightarrow 8 n = 48 \Rightarrow n = 6$

Page No 19.51:

Question 6:

If S_n denotes the sum of first n terms of an A.P. < a_n > such that $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}, then \frac{a_{m}}{a_{n}} =$
(a) $\frac{2 m + 1}{2 n + 1}$

(b) $\frac{2 m - 1}{2 n - 1}$

(c) $\frac{m - 1}{n - 1}$

(d) $\frac{m + 1}{n + 1}$

Answer:

(b) $\frac{2 m - 1}{2 n - 1}$

$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{\frac{m}{2} \{2 a + (m - 1) d\}}{\frac{n}{2} \{2 a + (n - 1) d\}} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{\{2 a + (m - 1) d\}}{\{2 a + (n - 1) d\}} = \frac{m}{n} \Rightarrow 2 a n + n d m - n d = 2 a m + n m d - m d \Rightarrow 2 a n - 2 a m - n d + m d = 0 \Rightarrow 2 a (n - m) - d (n - m) = 0 \Rightarrow 2 a (n - m) = d (n - m) \Rightarrow d = 2 a . . . . . (1)$

$Ratio of \frac{a_{m}}{a_{n}} = \frac{a + (m - 1) d}{a + (n - 1) d} \Rightarrow \frac{a_{m}}{a_{n}} = \frac{a + (m -) 2 a}{a + (n -) 2 a} From (1) = \frac{a + 2 a m - 2 a}{a + 2 a n - 2 a} = \frac{2 a m - a}{2 a n - a} = \frac{2 m - 1}{2 n - 1}$

Page No 19.51:

Question 7:

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
(a) 5
(b) 6
(c) 7
(d) 8

Answer:

(b) 6

$a = 1, a_{n} = 11, S_{n} = 36 ∵ a_{n} = 11 \Rightarrow a + (n - 1) d = 11 \Rightarrow 1 + (n - 1) d = 11 \Rightarrow (n - 1) d = 10 . . . . . (1) Also, S_{n} = 36 \Rightarrow \frac{n}{2} \{2 a + (n - 1) d\} = 36 \Rightarrow n \{2 + 10\} = 72 (Using (1)) \Rightarrow n = 6$

Page No 19.51:

Question 8:

If the sum of n terms of an A.P., is 3 n² + 5 n then which of its terms is 164?
(a) 26th
(b) 27th
(c) 28th
(d) none of these.

Answer:

(b) 27th

$S_{n} = 3 n^{2} + 5 n S_{1} = 3 {(1)}^{2} + 5 (1) = 8 ∴ a_{1} = 8 S_{2} = 3 {(2)}^{2} + 5 (2) = 22 ∴ a_{1} + a_{2} = 22 \Rightarrow a_{2} = 14 Common difference, d = 14 - 8 = 6 Also, a_{n} = 164 \Rightarrow a + (n - 1) d = 164 \Rightarrow 8 + (n - 1) 6 = 164 \Rightarrow n = 27$

Page No 19.51:

Question 9:

If the sum of n terms of an A.P. is 2 n² + 5 n, then its nth term is
(a) 4n − 3
(b) 3 n − 4
(c) 4 n + 3
(d) 3 n + 4

Answer:

(c) 4 n + 3

$S_{n} = 2 n^{2} + 5 n S_{1} = 2 . 1^{2} + 5.1 = 7 ∴ a_{1} = 7 S_{n} = 2 . 2^{2} + 5.2 = 18 ∴ a_{1} + a_{2} = 18 \Rightarrow a_{2} = 11 Common difference, d = 11 - 7 = 4 a_{n} = a + (n - 1) d = 7 + (n - 1) 4 = 4 n + 3$

Page No 19.51:

Question 10:

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series sin d [cosec a₁ cosec a₂ + cosec a₁ cosec a₃ + .... + cosec a_n_{− 1} cosec a_n] is
(a) sec a₁ − sec a_n
(b) cosec a₁ − cosec a_n
(c) cot a₁ − cot a_n
(d) tan a₁ − tan a_n

Answer:

(c) cot a₁ − cot a_{n

We have:
$\sin d (\cos e c a_{1} \cos e c a_{2} + \cos e c a_{2} \cos e c a_{3} + . . . . + \cos e c a_{n - 1} \cos e c a_{n}) = \frac{\sin d}{\sin a_{1} \sin a_{2}} + \frac{\sin d}{\sin a_{2} \sin a_{3}} + . . . . . + \frac{\sin d}{\sin a_{n - 1} \sin a_{n}} = \frac{\sin (a_{2} - a_{1})}{\sin a_{1} \sin a_{2}} + \frac{\sin (a_{3} - a_{2})}{\sin a_{2} \sin a_{3}} + . . . . + \frac{\sin (a_{n} - a_{n - 1})}{\sin a_{n - 1} \sin a_{n}} = \frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\sin a_{1} \sin a_{2}} + \frac{\sin a_{3} \cos a_{2} - \cos a_{3} \sin a_{2}}{\sin a_{1} \sin a_{2}} + . . . . . + \frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\sin a_{1} \sin a_{2}} = (\cot a_{1} - \cot a_{2}) + (\cot a_{2} - \cot a_{3}) + . . . . . + (\cot a_{n - 1} - \cot a_{n}) = \cot a_{1} - \cot a_{n}$}

Page No 19.51:

Question 11:

In the arithmetic progression whose common difference is non-zero, the sum of first 3 n terms is equal to the sum of next n terms. Then the ratio of the sum of the first 2 n terms to the next 2 n terms is
(a) 1/5
(b) 2/3
(c) 3/4
(d) none of these

Answer:

(a) 1/5

$S_{3 n} = S_{4 n} - S_{3 n} \Rightarrow 2 S_{3 n} = S_{4 n} \Rightarrow 2 \times \frac{3 n}{2} \{2 a + (3 n - 1) d\} = \frac{4 n}{2} \{2 a + (4 n - 1) d\} \Rightarrow 3 \{2 a + (3 n - 1) d\} = 2 \{2 a + (4 n - 1) d\} \Rightarrow 6 a + 9 n d - 3 d = 4 a + 8 n d - 2 d \Rightarrow 2 a + n d - d = 0 \Rightarrow 2 a + (n - 1) d = 0 . . . . (1)$

Required ratio: $\frac{S_{2 n}}{S_{4 n} - S_{2 n}}$

$\frac{S_{2 n}}{S_{4 n} - S_{2 n}} = \frac{\frac{2 n}{2} \{2 a + (2 n - 1) d\}}{\frac{4 n}{2} \{2 a + (4 n - 1) d\} - \frac{2 n}{2} \{2 a + (2 n - 1) d\}} = \frac{n (n d)}{2 n (3 n d) - n (n d)} = \frac{1}{6 - 1} = \frac{1}{5}$

Page No 19.51:

Question 12:

If a₁, a₂, a₃, .... a_n are in A.P. with common difference d, then the sum of the series sin d [sec a₁ sec a₂ + sec a₂ sec a₃ + .... + sec a_n_{− 1} sec a_n], is
(a) sec a₁ − sec a_n
(b) cosec a₁ − cosec a_n
(c) cot a₁ − cot a_n
(d) tan a_n − tan a₁

Answer:

(d) tan a_n − tan a₁
We have:

$\sin d (\sec a_{1} \sec a_{2} + \sec a_{2} \sec a_{3} + . . . . + \sec a_{n - 1} \sec a_{n}) = \frac{\sin d}{\cos a_{1} \cos a_{2}} + \frac{\sin d}{\cos a_{2} \cos a_{3}} + . . . . . + \frac{\sin d}{\cos a_{n - 1} \cos a_{n}} = \frac{\sin (a_{2} - a_{1})}{\cos a_{1} \cos a_{2}} + \frac{\sin (a_{3} - a_{2})}{\cos a_{2} \cos a_{3}} + . . . . + \frac{\sin (a_{n} - a_{n - 1})}{\cos a_{n - 1} \cos a_{n}} = \frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\cos a_{1} \cos a_{2}} + \frac{\sin a_{3} \cos a_{2} - \cos a_{3} \sin a_{2}}{\cos a_{1} \cos a_{2}} + . . . . . + \frac{\sin a_{2} \cos a_{1} - \cos a_{2} \sin a_{1}}{\cos a_{1} \cos a_{2}} = (\tan a_{1} - \tan a_{2}) + (\tan a_{2} - \tan a_{3}) + . . . . . + (\tan a_{n - 1} - \tan a_{n}) = \tan a_{1} - \tan a_{n}$

Page No 19.51:

Question 13:

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are
(a) 5, 10, 15, 20
(b) 4, 10, 16, 22
(c) 3, 7, 11, 15
(d) none of these

Answer:

(a) 5, 10, 15, 20

Let the four numbers in A.P. be as follows:
$a - 2 d, a - d, a, a + d$
Their sum = 50 (Given)
$\Rightarrow (a - 2 d) + (a - d) + a + (a + d) = 50 \Rightarrow 2 a - d = 25 . . . . . (1) Also, (a + d) = 4 (a - 2 d) \Rightarrow a + d = 4 a - 8 d \Rightarrow 3 d = a . . . . . . (2)$

From equations $(1) and (2),$ we get:
d = 5, a = 15

Hence, the numbers are $15 - 10, 15 - 5, 15, 15 + 5$ , i.e. 5, 10, 15, 20.

Page No 19.51:

Question 14:

If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29, then the value of n is
(a) 10
(b) 12
(c) 13
(d) 14

Answer:

(d) 14

The given series is 1, . . . . . . . . . . . , 31
There are n A.M.s between 1 and 31: $1, A_{1}, A_{2}, A_{3}, . . . . ., A_{n}, 31$ .

Common difference, d = $\frac{31 - 1}{n + 1} = \frac{30}{n + 1}$

Here, we have:
$\frac{A_{1}}{A_{n}} = \frac{3}{29} \Rightarrow \frac{1 + d}{1 + n d} = \frac{3}{29} \Rightarrow \frac{1 + \frac{30}{n + 1}}{1 + n \times \frac{30}{n + 1}} = \frac{3}{29} \Rightarrow \frac{n + 1 + 30}{n + 1 + 30 n} = \frac{3}{29} \Rightarrow \frac{n + 31}{31 n + 1} = \frac{3}{29} \Rightarrow 29 n + 899 = 93 n + 3 \Rightarrow 64 n = 896 \Rightarrow n = 14$

Page No 19.51:

Question 15:

Let S_n denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n − k S_n₋₁ + S_n_{− 2} , then k =
(a) 1
(b) 2
(c) 3
(d) none of these

Answer:

(b) 2

Let the A.P. be a, a+d, a+2d, a+3d...
Given:
$d = S_{n} - k S_{n - 1} + S_{n - 2}$
For n = 3, we have:

$d = (3 a + 3 d) - k (2 a + d) + a \Rightarrow 4 a + 2 d - k (2 a + d) = 0 \Rightarrow 2 (2 a + d) = k (2 a + d) \Rightarrow 2 = k$

Page No 19.51:

Question 16:

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2} - a^{2}}{k - (l + a)}$ , then k =
(a) S
(b) 2S
(c) 3S
(d) none of these

Answer:

(b) 2S

Given:

$S = \frac{n}{2} (l + a) \Rightarrow (l + a) = \frac{2 S}{n}$

$Also, d = \frac{l^{2} - a^{2}}{k - (l + a)} \Rightarrow d = \frac{(l + a) (l - a)}{k - (l + a)} \Rightarrow d = \frac{[(n - 1) d] \times \frac{2 S}{n}}{k - \frac{2 S}{n}} \Rightarrow k - \frac{2 S}{n} = (n - 1) \frac{2 S}{n} \Rightarrow k = \frac{2 S}{n} (n - 1 + 1) \Rightarrow k = 2 S$

Page No 19.51:

Question 17:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
(a) $\frac{1}{n}$

(b) $\frac{n - 1}{n}$

(c) $\frac{n + 1}{2 n}$

(d) $\frac{n + 1}{n}$

Answer:

(d) $\frac{n + 1}{n}$

Given:
Sum of the even natural numbers = k $\times$ Sum of the odd natural numbers
$\frac{n}{2} \{2 a + (n - 1) d\} = k \times \frac{n}{2} \{2 a + (n - 1) d\} \Rightarrow \{2 \times 2 + (n - 1) 2\} = k \times \{2 \times 1 + (n - 1) 2\} \Rightarrow \frac{4 + (n - 1) 2}{2 + (n - 1) 2} = k \Rightarrow \frac{n + 1}{n} = k$

Page No 19.52:

Question 18:

If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is
(a) $\frac{a b}{2 (b - a)}$

(b) $\frac{a b}{b - a}$

(c) $\frac{3 a b}{2 (b - a)}$

(d) none of these

Answer:

(c) $\frac{3 a b}{2 (b - a)}$

Let the A.P. be a, a+d, a+2d........a+nd.
Here, let d be the common difference and n be the total number of terms.

Given:
$a_{1} = a, a_{2} = b \Rightarrow a + d = b \Rightarrow d = b - a . . . . . (1) a_{n} = 2 a \Rightarrow a + (n - 1) d = 2 a \Rightarrow (n - 1) d = a \Rightarrow d = \frac{a}{n - 1} . . . . . (2)$

From equations $(1) and (2),$ we have:
$\Rightarrow \frac{a}{n - 1} = b - a \Rightarrow \frac{a}{b - a} + 1 = n \Rightarrow \frac{a + b - a}{b - a} = n \Rightarrow \frac{b}{b - a} = n$

Now, sum of n terms of an A.P.:
$S = \frac{n}{2} \{a + a_{n}\} = \frac{n}{2} (3 a) = \frac{3 a b}{2 (b - a)}$

Page No 19.52:

Question 19:

If, S₁ is the sum of an arithmetic progression of 'n' odd number of terms and S₂ the sum of the terms of the series in odd places, then $\frac{S_{1}}{S_{2}}$ =
(a) $\frac{2 n}{n + 1}$

(b) $\frac{n}{n + 1}$

(c) $\frac{n + 1}{2 n}$

(d) $\frac{n + 1}{n}$

Answer:

(a) $\frac{2 n}{n + 1}$

Let n be an odd number.
Given:

$S_{1} = Sum of odd number of terms = \frac{n}{2} \{2 a + (n - 1) d\} . . . . . (1) Since n is odd, the number of odd places = \frac{n + 1}{2} S_{2} = Sum of the terms of a series in odd places = \frac{(\frac{n + 1}{2})}{2} \{2 a + (\frac{n + 1}{2} - 1) 2 d\} = \frac{n + 1}{4} \{2 a + (n - 1) d\} . . . . . (2)$

From equations $(1) and (2)$ , we have:

$\frac{S_{1}}{S_{2}} = \frac{\frac{n}{2} \{2 a + (n - 1) d\}}{\frac{n + 1}{4} \{2 a + (n - 1) d\}} = \frac{2 n}{n + 1}$

Page No 19.52:

Question 20:

If in an A.P., S_n = n²p and S_m = m²p, where S_r denotes the sum of r terms of the A.P., then S_p is equal to
(a) $\frac{1}{2} p^{3}$

(b) mn p

(c) P³

(d) (m + n) p²

Answer:

(c) p³

Given:
$S_{n} = n^{2} p \Rightarrow \frac{n}{2} \{2 a + (n - 1) d\} = n^{2} p \Rightarrow 2 a + (n - 1) d = 2 n p \Rightarrow 2 a = 2 n p - (n - 1) d . . . . . (1) S_{m} = m^{2} p \Rightarrow \frac{m}{2} \{2 a + (m - 1) d\} = m^{2} p \Rightarrow 2 a + (m - 1) d = 2 m p \Rightarrow 2 a = 2 m p - (m - 1) d . . . . . (2)$

From $(1) and (2)$ , we have:

$2 n p - (n - 1) d = 2 m p - (m - 1) d \Rightarrow 2 p (n - m) = d (n - 1 - m + 1) \Rightarrow 2 p = d$

Substituting d = 2p in equation $(1)$ , we get:
a = p

Sum of p terms of the A.P. is given by:

$\frac{p}{2} \{2 a + (p - 1) d\} = \frac{p}{2} \{2 p + (p - 1) 2 p\} = p^{3}$

Page No 19.52:

Question 21:

Mark the correct alternative in the following question:

If in an A.P., the pth term is q and (p + q)th term is zero, then the qth term is

(a) $-$ p (b) p (c) p + q (d) p $-$ q

Answer:

$As, a_{p} = q \Rightarrow a + (p - 1) d = q . . . . . (i) Also, a_{(p + q)} = 0 \Rightarrow a + (p + q - 1) d = 0 . . . . . (ii) Subtracting (i) from (ii), we get a + (p + q - 1) d - a - (p - 1) d = 0 - q \Rightarrow (p + q - 1 - p + 1) d = - q \Rightarrow q d = - q \Rightarrow d = \frac{- q}{q} \Rightarrow d = - 1 Substituting d = - 1 in (i), we get a + (p - 1) \times (- 1) = q \Rightarrow a - p + 1 = q \Rightarrow a = p + q - 1 Now, a_{q} = a + (q - 1) d = p + q - 1 + (q - 1) \times (- 1) = p + q - 1 - q + 1 = p$

Hence, the correct alternative is option (b).

Page No 19.52:

Question 22:

Mark the correct alternative in the following question:

The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is

(a) 191 (b) 193 (c) 211 (d) none of these

Answer:

$As, the common difference of the A . P . 3, 7, 11, 15, . . . = 7 - 3 = 4 and the common difference of the A . P . 1, 6, 11, 16, . . . = 6 - 1 = 5 And, the common terms of both the A . P . s will be in A . P . So, the common difference of the A . P . of the common terms, d = LCM (4, 5) = 4 \times 5 = 20 and its first common term, a = 11 Now, the tenth common term, a_{10} = a + (10 - 1) d = 11 + 9 \times 20 = 11 + 180 = 191$

Hence, the correct alternative is option (a).

Page No 19.52:

Question 23:

Mark the correct alternative in the following question:

$If in an A . P . S_{n} = n^{2} q and S_{m} = m^{2} q, where S_{r} denotes the sum of r terms of the A . P ., then S_{q} equals (a) \frac{q^{3}}{2} (b) m n q (c) q^{3} (d) (m^{2} + n^{2}) q$

Answer:

$As, S_{n} = n^{2} q \Rightarrow \frac{n}{2} [2 a + (n - 1) d] = n^{2} q \Rightarrow 2 a + (n - 1) d = \frac{n^{2} q \times 2}{n} \Rightarrow 2 a + (n - 1) d = 2 n q . . . . . (i) Also, S_{m} = m^{2} q \Rightarrow \frac{m}{2} [2 a + (m - 1) d] = m^{2} q \Rightarrow 2 a + (m - 1) d = \frac{m^{2} q \times 2}{m} \Rightarrow 2 a + (m - 1) d = 2 m q . . . . . (ii) Subtracting (i) from (ii), we get 2 a + (n - 1) d - 2 a - (m - 1) d = 2 n q - 2 m q \Rightarrow (n - 1 - m + 1) d = 2 q (n - m) \Rightarrow (n - m) d = 2 q (n - m) \Rightarrow d = \frac{2 q (n - m)}{(n - m)} \Rightarrow d = 2 q Substituting d = 2 q in (ii), we get 2 a + (m - 1) 2 q = 2 m q \Rightarrow 2 a + 2 m q - 2 q = 2 m q \Rightarrow 2 a = 2 q \Rightarrow a = q Now, S_{q} = \frac{q}{2} [2 a + (q - 1) d] = \frac{q}{2} [2 q + (q - 1) 2 q] = \frac{q}{2} [2 q + 2 q^{2} - 2 q] = \frac{q}{2} [2 q^{2}] = q^{3}$

Hence, the correct alternative is option (c).

Page No 19.52:

Question 24:

Mark the correct alternative in the following question:

Let S_n denote the sum of first n terms of an A.P. If S₂_n = 3S_n, then S₃_n : S_n is equal to

(a) 4 (b) 6 (c) 8 (d) 10

Answer:

$As, S_{2 n} = 3 S_{n} \Rightarrow \frac{2 n}{2} [2 a + (2 n - 1) d] = \frac{3 n}{2} [2 a + (n - 1) d] \Rightarrow 2 [2 a + (2 n - 1) d] = 3 [2 a + (n - 1) d] \Rightarrow 4 a + 2 (2 n - 1) d = 6 a + 3 (n - 1) d \Rightarrow 4 a + 4 n d - 2 d = 6 a + 3 n d - 3 d \Rightarrow 6 a - 4 a + 3 n d - 3 d - 4 n d + 2 d = 0 \Rightarrow 2 a - n d - d = 0 \Rightarrow 2 a - (n + 1) d = 0 \Rightarrow 2 a = (n + 1) d . . . . . (i) Now, \frac{S_{3 n}}{S_{n}} = \frac{\frac{3 n}{2} [2 a + (3 n - 1) d]}{\frac{n}{2} [2 a + (n - 1) d]} = \frac{3 [(n + 1) d + (3 n - 1) d]}{[(n + 1) d + (n - 1) d]} [Using (i)] = \frac{3 [n d + d + 3 n d - d]}{[n d + d + n d - d]} = \frac{3 [4 n d]}{[2 n d]} = 3 \times 2 = 6$

Hence, the correct alternative is option (b).

Page No 19.52:

Question 25:

If the sum of n terms of an A.P. is given by S_n = 3n + 2n², then the common difference of the A.P. is
(a) 3
(b) 2
(c) 6
(d) 4

Answer:

Let S_n denote sum of n terms of an A.P
Such that S_n = 3n + 2n²
i.e S₁ = 3(1) + 2(1)²
S₁ = 3 + 2 = 5
where S₁ = a (first term only)
S₂ = a + (a + d) where a + d represents second term and d is common difference.
$\begin{array}{rcl} S_{2} & = & 3 (2) + 2 {(2)}^{2} \\ = & 6 + 2 (4) \\ = & 6 + 8 \\ S_{2} & = & 14 \end{array}$
i.e 2a + d = 14
i.e 10 + d = 14
d = 4
i.e common difference of A.P is 4

Hence, the correct answer is option D.

Page No 19.52:

Question 26:

If 9 times the 9^th term of an A.P. is equal to 13 times the 13th term, then the 22^nd term of the A.P. is
(a) 0
(b) 22
(c) 220
(d) 198

Answer:

Let a denote the first term of A.P and d denote the common difference of the A.P
Then ninth term is a + 8d and 13^th term is a + 12d
According to given condition,
9(a + 8d) = 13(a + 12d)
i.e 9a + 72d = 13a + 156d
i.e –156d + 72d = 4a
i.e –84d = 4a
i.e a = – 21d
i.e a + 21d = 0
which represents 22^nd term of A.P.
∴ The 22^ndterm of the A.P is 0
Hence, the correct answer is option A.

Page No 19.52:

Question 1:

The sum of the terms equidistant from the beginning and end in an A.P. is always same and is equal to the sum of ________ and ________ terms.

Answer:

Let a₁, a₂, ............ a_nbe an A.P with common difference d.
Then k^th term is a_k = a₁ + (k – 1)d and k^th term from last is the (n – k + 1)^th term from beginning
$i . e a_{n - k + 1} = a_{1} + (n - k + 1 - 1) d a_{n - k + 1} = a_{1} + (n - k) d$
∴ k^thterm from beginning + k^thterm from the end is
$a_{1} + (k - 1) d + a_{1} + (n - k) d = 2 a_{1} + d (k - 1 + n - k) d = 2 a_{1} + (n - 1) d = a_{1} + (a_{1} + (n - 1) d) = a_{1} + a_{n}$
∴ Sum of the terms equidistant from the beginning of an A.P and end of A.P is always same and is equal to sum of first term and n^th term or last term of the A.P.

Page No 19.52:

Question 2:

The minimum value of 4^x + 4^{1 – x}, x ∈ R, is ___________.

Answer:

Since
Arithmetic mean ≥ geometric mean of 4^x and 4^{1 – x}
$i . e \frac{4^{x} + 4^{1 - x}}{2} \geq \sqrt{4^{x} . 4^{1 - x}} i . e \frac{4^{x} + 4^{1 - x}}{2} \geq \sqrt{4^{x} . 4 . 4^{- x}} i . e \frac{4^{x} + 4^{1 - x}}{2} \geq 2 i . e 4^{x} + 4^{1 - x} \geq 2 \times 2 = 4$

i.e minimum value of 4^x + 4^{1 – x} is 4.

Page No 19.53:

Question 3:

If the first, second and last terms of an A.P. are a, b and 2a respectively, then the sum of its terms is ___________.

Answer:

Let first term of A.P be a
Second term of A.P be b and last term of A.P be given by 2a
Since d = a₂ – a₁= b – a

$\begin{array}{rcl} n^{th} term i . e a_{n} & = & a + (n - 1) d \\ = & a + (n - 1) (b - a) = 2 a (given) \end{array}$
$\Rightarrow a + (n - 1) (b - a) = 2 a \Rightarrow (n - 1) (b - a) = a i . e n - 1 = \frac{a}{b - a} i . e n = \frac{a}{b - a} + 1 = \frac{b}{b - a}$
∴ S_ni.e sum of n terms of A.P is
$\frac{n}{2} (a_{1} + a_{n}) = \frac{n}{2} (a + 2 a) = \frac{b}{2 (b - a)} (3 a)$
∴ Sum of its term = $\frac{3 a b}{2 (b - a)}$

Page No 19.53:

Question 4:

The number of terms in an A.P. whose first term is 10, last term is 50 and the sum of all terms is 300, is ___________.

Answer:

First term of A.P is given = 10
Last term of A.P is given = 50
Sum of all terms S_n = 300 (given)
Since sum of all terms $S_{n} = \frac{n (First term + Last term)}{2}$
$i . e 300 = n {(\frac{10 + 50}{2})}^{2} i . e 300 = n (\frac{60}{2}) i . e 300 = n (30) i . e n = 10$
Hence, number of terms of an A.P is 10.

Page No 19.53:

Question 5:

The arithmetic mean of first n natural numbers is ___________.

Answer:

First n natural numbers are 1, 2, 3, ...........n
Sum of these numbers is 1 + 2 + 3 + ........... + n $= \frac{n (n + 1)}{2}$

$\begin{array}{rcl} Hence arithmetic mean of first n natural numbers & = & \frac{Sum of numbers}{Total numbers} \\ = & \frac{n (n + 1) / 2}{n} \end{array}$
i.e Arithmetic mean of first n natural number = $\frac{n + 1}{2}$

Page No 19.53:

Question 6:

The sum of first n odd natural numbers is ___________.

Answer:

First n odd natural numbers are 1, 3, 5, 7, ............. n
Here first term a = 1
Common difference d = 2
Sum of A.P is
$\frac{n}{2} \{2 a + (n - 1) d\} = \frac{n}{2} \{2 (2) + (n - 1) 2\} = \frac{n}{2} \{4 + 2 n - 2\} = \frac{n}{2} \{2 + 2 n\}$
i.e Sum of first n even numbers = n(n + 1).

Page No 19.53:

Question 7:

The sum of first n even natural numbers is ___________.

Answer:

Ans

Page No 19.53:

Question 8:

If n is even, then the sum of first n terms of the series 1 – 2 + 3 – 4 + 5 – 6 +..., is ___________.

Answer:

For even number n,
Consider 1 – 2 + 3 – 4 + 5 – 6 + ................
for 1 – 2 = – 1
3 – 4 = – 1
5 – 6 = –1
i.e Combining two terms at a time given –1
So, if in all, n even terms are there
Sum of the expression 1 – 2 + 3 – 4 + .......... n – 1 – n.
$= - \frac{n}{2}$

Page No 19.53:

Question 9:

If twice the 11^th term of an A.P is equal to 7 times of its 21^st terms, then the value of 25^th term is ___________.

Answer:

Let us suppose a denote the first term of A.P and d represents the common difference
Then 11^th term is a + 10d
21^st term is a + 20d
According to given condition,
2(11^th term) = 7(21^st term)
i.e 2(a + 10d) = 7(a + 20d)
i.e 2a + 20d = 7a + 140d
i. 5a + 120d = 0
i.e 5(a + 24d) = 0
Since 5 ≠ 0
⇒ a + 24d = 0
i.e 25^th term of A.P is 0.

Page No 19.53:

Question 10:

If the sums of n terms of two arithmetic progressions are the ratio (2n + 3) : (6n + 5), then the ratio of their 13^th terms is ___________.

Answer:

Let a₁ and d₂ represent the first term and common difference of first arithmetic progression.
Also, a₂ and d₂ represent the first term and common difference of second arithmetic progression then sum of n terms of first A.P
${(S_{n})}_{1} = \frac{n}{2} \{2 a_{1} + (n - 1) d_{1}\}$
and sum of n terms of second A.P
${(S_{n})}_{2} = \frac{n}{2} \{2 a_{2} + (n - 1) d_{2}\}$
Now, according to given condition
$\frac{{(S_{n})}_{1}}{{(S_{n})}_{2}} = \frac{2 n + 3}{6 n + 5} i . e \frac{2 n + 3}{6 n + 5} = \frac{\frac{n}{2} \{2 a_{1} + (n - 1) d_{1}\}}{\frac{n}{2} \{2 a_{2} + (n - 1) d_{2}\}} \frac{2 n + 3}{6 n + 5} = \frac{a_{1} + \frac{(n - 1)}{2} d_{1}}{a_{2} + \frac{(n - 1)}{2} d_{2}}$
Since n^thterm of any A.P is a + (n – 1)d
∴ 13^th term of first A.P is a₁ + 12d₁ and 13^th term of second A.P is a₂ + 12d₂
$i . e \frac{n - 1}{2} = 12 i . e \frac{n - 1}{2} = 12 i . e n = 24 + 1 i . e n = 25 ∴ \frac{a_{1} + 12 d_{1}}{a_{2} + 12 d_{2}} = (\frac{2 n + 3}{6 n + 5}) a t n = 25 = \frac{2 (25) + 3}{6 (25) + 5} ∴ \frac{{(a_{13})}_{1}}{{(a_{13})}_{2}} = \frac{53}{155}$
i.e ratio of their 13^th terms is $\frac{53}{155}$ .

Page No 19.53:

Question 11:

The sum of n arithmetic means between a and b is ___________.

Answer:

Let a and b be two given numbers
Let A₁, A₂, ______ A_n be arithmetic means between a and b.
∴ a, A₁, A₂, _____ A_n , b are in A.P
Here total number of terms is n + 2 first term is a last term is b.
∴ Sum of first n + 2 terms = $\frac{(n + 2)}{2} \{a + b\}$
$i . e a + A_{1} + A_{2} + . . . . . . . . . . + A_{n} + b = (\frac{n + 2}{2}) (a + b) i . e A_{1} + A_{2} + . . . . . . . . + A_{n} = \frac{(n + 2) (a + b)}{2} - a - b = \frac{(n + 2) (a + b)}{2} - (a + b) = (a + b) \{\frac{n + 2}{2} - 1\} = (a + b) \{\frac{n + 2 - 2}{2}\} = (a + b) \frac{n}{2} i . e A_{1} + A_{2} + . . . . . . . . + A_{n} = n (\frac{a + b}{2})$
i.e

Page No 19.53:

Question 12:

If the sum of n arithmetic means between 9 and 51 is 270, then the value of n is ___________.

Answer:

Let a = 9
b = 51
Given sum of n arithmetic means between 9 and 51 is 270.
Let A₁, A₂, ______ A_n be arithmetic means between a and b.
∴ a, A₁, A₂, _____ A_n , b are in A.P
Here total number of terms is n + 2 first term is a last term is b.
∴ Sum of first n + 2 terms = $\frac{(n + 2)}{2} \{a + b\}$
$A_{1} + A_{2} + . . . . . . . . . . . + A_{n} = \frac{n (a + b)}{2} i . e 270 = n (\frac{9 + 51}{2}) i . e 270 = n (\frac{60}{2}) i . e \frac{270}{30} = n i . e n = 9$
Hence, value of n is 9.

Page No 19.53:

Question 13:

The sum of 4 arithmetic means between 3 and 23 is ___________.

Answer:

Let a = 3
b = 23
To find sum of 4 arithmetic term ..............
here n = 4
∴ Sum of 4 arithmetic terms is means between 3 and 23
$4 (\frac{a + b}{2}) = 4 (\frac{3 + 23}{2}) = 2 (26)$
i.e sum of 4 arithmetic means between 3 and 23 = 52.

Page No 19.53:

Question 14:

If $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the A.M. of a and b, then n = ___________.

Answer:

Given:- $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is arithmetic mean of a and b.
Since arithmetic mean between a and b is $\frac{a + b}{2}$
$\Rightarrow \frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}} = \frac{a + b}{2} 2 (a^{n + 1} + b^{n + 1}) = (a + b) (a^{n} + b^{n}) i . e 2 a^{n + 1} + 2 b^{n + 1} = a^{n + 1} + a b^{n} + b a^{n} + b^{n + 1} i . e a^{n + 1} + b^{n + 1} = a b^{n} + b a^{n} i . e a^{n + 1} - a^{n} b = a b^{n} - b^{n + 1} i . e a^{n} (a - b) = b^{n} (a - b) i . e a^{n} = b^{n} (Since a - b \neq 0 a a n d b are distinct) i . e {(\frac{a}{b})}^{n} = 1 i . e n = 0$

Page No 19.53:

Question 1:

Write the common difference of an A.P. whose nth term is xn + y.

Answer:

$We have : a_{n} = x n + y ∴ a_{1} = x + y a_{2} = 2 x + y$
Common difference of an A.P., d = $a_{2} - a_{1}$
$\Rightarrow (2 x + y) - (x + y) \Rightarrow x$

Page No 19.53:

Question 2:

Write the common difference of an A.P. the sum of whose first n terms is $\frac{p}{2} n^{2} + Q n$ .

Answer:

Sum of the first n terms of an A.P. = $\frac{p}{2} n^{2} + Q n$
Sum of one term of an A.P. = $S_{1}$
$\Rightarrow \frac{p}{2} {(1)}^{2} + Q (1) \Rightarrow \frac{p}{2} + Q$

Sum of two terms of an A.P. = $S_{2}$
$\Rightarrow \frac{p}{2} {(2)}^{2} + Q (2) \Rightarrow 2 p + 2 Q$
Now, we have:
$a_{1} + a_{2} = S_{2} \Rightarrow \frac{p}{2} + Q + a_{2} = 2 p + 2 Q \Rightarrow a_{2} = Q + \frac{3}{2} p$

Common difference:
$d = a_{2} - a_{1} = (Q + \frac{3}{2} p) - (Q + \frac{p}{2}) = p$

Page No 19.53:

Question 3:

If the sum of n terms of an AP is 2n² + 3n, then write its nth term.

Answer:

Given:
$S_{n} = 2 n^{2} + 3 n$
$\Rightarrow S_{1} = 2 {(1)}^{2} + 3 (1) = 5 S_{2} = 2 {(2)}^{2} + 3 (2) = 14 ∴ a_{1} + a_{2} = 14 \Rightarrow 5 + a_{2} = 14 \Rightarrow a_{2} = 9$

Common difference, d  = $a_{2} - a_{1}$
                                    = 9 $-$ 5
                                      = 4
nth term = a + $(n - 1) d$
               = 5+ $(n - 1)$ 4
               = 4n+1

Page No 19.53:

Question 4:

If log 2, log (2^x − 1) and log (2^x + 3) are in A.P., write the value of x.

Answer:

The numbers log 2, log (2^x − 1) and log (2^x + 3) are in A.P.

$∴ \log (2^{x} - 1) - \log 2 = \log (2^{x} + 3) - \log (2^{x} - 1) \Rightarrow \log (\frac{2^{x} - 1}{2}) = \log (\frac{2^{x} + 3}{2^{x} - 1}) \Rightarrow \frac{2^{x} - 1}{2} = \frac{2^{x} + 3}{2^{x} - 1} \Rightarrow {(2^{x} - 1)}^{2} = 2 (2^{x} + 3) \Rightarrow 2^{2 x} + 1 - 2 . 2^{x} = 2 . 2^{x} + 6 \Rightarrow 2^{2 x} - 4 . 2^{x} - 5 = 0$

$Let 2^{x} = y . \Rightarrow y^{2} - 4 y - 5 = 0 \Rightarrow (y - 5) (y + 1) = 0 \Rightarrow y = 5 or y = - 1 ∴ 2^{x} = 5 or 2^{x} = - 1 (not possible) Taking \log on both the sides, we get : \log 2^{x} = \log 5 \Rightarrow x \log 2 = \log 5 \Rightarrow x = \frac{\log 5}{\log 2} = \log_{2} 5 \Rightarrow x = \log_{2} 5$

Disclaimer: The question in the book has some error, so, this solution is not matching with the solution given in the book. This solution here is created according to the question given in the book.

Page No 19.53:

Question 5:

If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their mth terms.

Answer:

Given:

$\frac{S_{n}}{{S_{n}}^{1}} = \frac{2 n + 5}{3 n + 4} \Rightarrow \frac{\frac{n}{2} \{2 a + (n - 1) d\}}{\frac{n}{2} \{2 b + (n - 1) d^{1}\}} = \frac{2 n + 5}{3 n + 4} \Rightarrow \frac{2 a + (n - 1) d}{2 b + (n - 1) d^{1}} = \frac{2 n + 5}{3 n + 4} . . . (1)$

Ratio of their m terms = $\frac{a_{m}}{b_{m}}$
To find the ratio of the m^th terms, replace n by 2m $-$ 1 in equation (1).
$\Rightarrow \frac{2 a + (2 m - 2) d}{2 b + (2 m - 2) d^{1}} = \frac{2 (2 m - 1) + 5}{3 (2 m - 1) + 4} \Rightarrow \frac{a + (m - 1) d}{b + (m - 1) d^{1}} = \frac{4 m - 2 + 3}{6 m - 3 + 4} \Rightarrow \frac{a_{m}}{b_{m}} = \frac{4 m + 1}{6 m + 1}$

Page No 19.53:

Question 6:

Write the sum of first n odd natural numbers.

Answer:

We need to find the sum of 1, 3, 5, 7... upto n terms.
Here, a = 1, d = 2

We know:
$S_{n} = \frac{n}{2} \{2 a + (n - 1) d\} = \frac{n}{2} \{2 \times 1 + (n - 1) 2\} = n^{2}$

Therefore, the sum of the first n odd numbers is $n^{2}$ .

Page No 19.53:

Question 7:

Write the sum of first n even natural numbers.

Answer:

We need to find the sum of 2, 4, 6, 8...upto n terms.
Here, a = 2, d = 2

We know:
$S_{n} = \frac{n}{2} \{2 a + (n - 1) d\} = \frac{n}{2} \{2 \times 2 + (n - 1) 2\} = n (n + 1)$

Therefore, the sum of the first n odd numbers is $n (n + 1)$ .

Page No 19.53:

Question 8:

Write the value of n for which nth terms of the A.P.s 3, 10, 17, ... and 63, 65, 67, .... are equal.

Answer:

For the first series, a = 3, $d_{1}$ = 7
For the second series, b = 63, $d_{2}$ = 2

Given:
$a_{n} = b_{n} \Rightarrow a + (n - 1) d_{1} = b + (n - 1) d_{2} \Rightarrow 3 + (n - 1) 7 = 63 + (n - 1) 2 \Rightarrow 3 + 7 n - 7 = 63 + 2 n - 2 \Rightarrow 5 n = 65 \Rightarrow n = 13$

Hence, the 13th terms of both the series are the same.

Page No 19.53:

Question 9:

If $\frac{3 + 5 + 7 + . . . + upto n terms}{5 + 8 + 11 + . . . . upto 10 terms}$ 7, then find the value of n.

Answer:

$\frac{3 + 5 + 7 + . . . + upto n terms}{5 + 8 + 11 + . . . . upto 10 terms} = 7 \Rightarrow \frac{\frac{n}{2} \{2 \times 3 + (n - 1) 2\}}{\frac{10}{2} \{2 \times 5 + (10 - 1) 3\}} = 7 \Rightarrow \frac{n (4 + 2 n)}{370} = 7 \Rightarrow n^{2} + 2 n - 1295 = 0 \Rightarrow n^{2} + 37 n - 35 n - 1295 = 0 \Rightarrow (n + 37) (n - 35) \Rightarrow n = 35, n = - 37 Rejecting the negative value, we get : \Rightarrow n = 35$

Page No 19.54:

Question 10:

If mth term of an A.P. is n and nth term is m, then write its pth term.

Answer:

Given:

$a_{m} = n \Rightarrow a + (m - 1) d = n . . . . (1) a_{n} = m \Rightarrow a + (n - 1) d = m . . . . (2)$

Solving equations $(1) and (2)$ , we get:
d = $- 1$
a = n+m $- 1$

pth term:
$a_{p} = a + (p - 1) d = n + m - 1 + (p - 1) (- 1) = n + m - p$

Hence, the pth term is n + m $-$ p.

Page No 19.54:

Question 11:

If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.

Answer:

Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.

Now,

$\frac{S_{n}}{S_{n}'} = \frac{(3 n + 2)}{(2 n + 3)} \Rightarrow \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 a' + (n - 1) d']} = \frac{(3 n + 2)}{(2 n + 3)} \Rightarrow \frac{[2 a + (n - 1) d]}{[2 a' + (n - 1) d']} = \frac{(3 n + 2)}{(2 n + 3)} Let n = 23 \Rightarrow \frac{2 a + (23 - 1) d}{2 a' + (23 - 1) d'} = \frac{3 \times 23 + 2}{2 \times 23 + 3} \Rightarrow \frac{2 a + 22 d}{2 a' + 22 d'} = \frac{69 + 2}{46 + 3} \Rightarrow \frac{2 (a + 11 d)}{2 (a' + 11 d')} = \frac{71}{49} ∴ \frac{a_{12}}{a_{12}'} = \frac{71}{49}$

So, the ratio of their 12th terms is 71 : 49.

View NCERT Solutions for all chapters of Class 11