Mathematics Part i Solutions Solutions for Class 10 Maths Chapter 3 - Arithmetic Progression

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Page No 61:

Question 1:

Which of the following sequences are A.P. ? If they are A.P. find the common difference .
(1) 2, 4, 6, 8,....
(2) 2, $\frac{5}{2}$ , 3, $\frac{7}{3}$ ,....
(3) –10, –6, –2, 2...
(4) 0.3, 0.33, 0.333,...
(5) 0, –4, –8, –12,...
(6) $- \frac{1}{5}, - \frac{1}{5}, - \frac{1}{5}, . . .$
(7) 3, $3 + \sqrt{2}, 3 + 2 \sqrt{2}, 3 + 3 \sqrt{2}$
(8) 127, 132, 137,...

Answer:

(1) 2, 4, 6, 8,....

The given sequence is an A.P.
Common difference = Second term − First term
                                 = 4 − 2
                                 = 2

(2) 2, $\frac{5}{2}$ , 3, $\frac{7}{3}$ ,....

The given sequence is not an A.P.

(3) –10, –6, –2, 2...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (−6) − (−10)
                                 = 4

(4) 0.3, 0.33, 0.333,...

The given sequence is not an A.P.

(5) 0, –4, –8, –12,...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (−4) − (0)
                                 = −4

(6) $- \frac{1}{5}, - \frac{1}{5}, - \frac{1}{5}, . . .$

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (− $\frac{1}{5}$ ) − (− $\frac{1}{5}$ )
                                 = 0

(7) 3, $3 + \sqrt{2}, 3 + 2 \sqrt{2}, 3 + 3 \sqrt{2}$

The given sequence is an A.P.
Common difference = Second term − First term
                                 = $3 + \sqrt{2} - 3$
                                 = $\sqrt{2}$

(8) 127, 132, 137,...

The given sequence is an A.P.
Common difference = Second term − First term
                                 = (132) − (127)
                                 = 5

Page No 61:

Question 2:

Write an A.P. whose first term is a and common difference is d in each of the following.
(1) a = 10, d = 5
(2) a = –3, d = 0
(3) a = –7, d = $\frac{1}{2}$
(4) a = –1.25, d = 3
(5) a = 6, d = –3
(6) a = 19, d = –4

Answer:

(1) a = 10, d = 5

First term = a = 10
Second term = a + d = 10 + 5 = 15
Third term = a + 2d = 10 + 10 = 20
and so on...

Hence, the required A.P. is 10, 15, 20, 25, ...

(2) a = –3, d = 0

First term = a = –3
Second term = a + d = –3 + 0 = –3
Third term = a + 2d = –3 + 0 = –3
and so on...

Hence, the required A.P. is –3, –3, –3, ...

(3) a = –7, d = $\frac{1}{2}$

First term = a = –7
Second term = a + d = $- 7 + \frac{1}{2} = - \frac{13}{2}$
Third term = a + 2d = –7 + 1 = –6
and so on...

Hence, the required A.P. is –7, – $\frac{13}{2}$ , –6, ...

(4) a = –1.25, d = 3

First term = a = –1.25
Second term = a + d = –1.25 + 3 = 1.75
Third term = a + 2d = –1.25 + 6 = 4.75
and so on...

Hence, the required A.P. is –1.25, 1.75, 4.75, ...

(5) a = 6, d = –3

First term = a = 6
Second term = a + d = 6 – 3 = 3
Third term = a + 2d = 6 – 6 = 0
and so on...

Hence, the required A.P. is 6, 3, 0, ...

(6) a = 19, d = –4

First term = a = 19
Second term = a + d = 19 – 4 = 15
Third term = a + 2d = 19 – 8 = 11
and so on...

Hence, the required A.P. is 19, 15, 11, ...

Page No 62:

Question 3:

Find the first term and common difference for each of the A.P.
(1) 5, 1, –3, –7,...
(2) 0.6, 0.9, 1.2,1.5,...
(3) 127, 135, 143, 151,...
(4) $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, . . . .$

Answer:

(1) 5, 1, –3, –7,...

First term = 5
Common difference = Second term – First term
                                 = 1 – 5
                                 = –4

(2) 0.6, 0.9, 1.2,1.5,...

First term = 0.6
Common difference = Second term – First term
                                 = 0.9 – 0.6
                                 = 0.3

(3) 127, 135, 143, 151,...

First term = 127
Common difference = Second term – First term
                                 = 135 – 127
                                 = 8

(4) $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, . . . .$

First term = $\frac{1}{4}$
Common difference = Second term – First term
                                 = $\frac{3}{4} - \frac{1}{4}$
                                 = $\frac{1}{2}$

Page No 66:

Question 1:

Write the correct number in the given boxes from the following A. P.
(i) 1, 8, 15, 22,...
Here a = $x$ , t₁= $x$ , t₂ = $x$ , t₃ = $x$ ,
t₂ – t₁ = $x$ – $x$ = $x$
t₃ – t₂ = $x$ – $x$ = $x$ ∴ d = $x$

(ii) 3, 6, 9, 12,...
Here t₁= $x$ , t₂ = $x$ , t₃ = $x$ , t₄ = $x$ ,
t₂ – t₁ = $x$ , t₃ – t₂ = $x$ ∴ d = $x$

(iii) –3, –8, –13, –18,...
Here t₃= $x$ , t₂ = $x$ , t₄ = $x$ , t₁ = $x$ ,
t₂ – t₁ = $x$ , t₃ – t₂ = $x$ ∴ a = $x$ , d = $x$

(iv) 70, 60, 50, 40,...
Here t₁= $x$ , t₂ = $x$ , t₃ = $x$ ,...
∴ a = $x$ , d = $x$

Answer:

(i) 1, 8, 15, 22,...
Here a = $1$ , t₁= $1$ , t₂ = $8$ , t₃ = $15$ ,
t₂ – t₁ = $8$ – $1$ = $7$
t₃ – t₂ = $15$ – $8$ = $7$ ∴ d = $7$

(ii) 3, 6, 9, 12,...
Here t₁= $3$ , t₂ = $6$ , t₃ = $9$ , t₄ = $12$ ,
t₂ – t₁ = $6 - 3 = 3 x$ , t₃ – t₂ = $9 - 6 = 3$ ∴ d = $3$

(iii) –3, –8, –13, –18,...
Here t₃= $- 13$ , t₂ = $- 8 x$ , t₄ = $- 18 x$ , t₁ = $- 3 x$ ,
t₂ – t₁ = $- 8 - (- 3) = - 5 x$ , t₃ – t₂ = $- 13 - (- 8) = - 5 x$ ∴ a = $- 3$ , d = $- 5 x$

(iv) 70, 60, 50, 40,...
Here t₁= $70$ , t₂ = $60$ , t₃ = $50$ ,...
∴ a = $70$ , d = $- 10 x$

Page No 66:

Question 2:

Decide whether following sequence is an A.P., if so find the 20^th term of the progression.
–12, –5, 2, 9, 16, 23, 30,...

Answer:

The given sequence is –12, –5, 2, 9, 16, 23, 30,...

Here,
First term (a) = a₁ = –12
Second term = a₂ = –5
Third term = a₃ = 2

Common difference (d) = a₂ – a₁ = –5 – (–12) = 7
= a₃ – a₂ = 2 – (–5) = 7

Since, a₂ – a₁ = a₃ – a₂

Hence, the given sequence is an A.P.

Now,
$a_{20} = a + (n - 1) d = - 12 + (20 - 1) 7 = - 12 + (19) 7 = 121$

Hence, the 20^th term of the progression is 121.

Page No 66:

Question 3:

Given Arithmetic Progression 12, 16, 20, 24, . . . Find the 24^th term of this progression.

Answer:

The given sequence is 12, 16, 20, 24, . . .

Here,
First term (a) = 12

Common difference (d) = a₂ – a₁ = 16 – (12) = 4

Now,
$a_{24} = a + (n - 1) d = 12 + (24 - 1) 4 = 12 + (23) 4 = 104$

Hence, the 24^th term of the progression is 104.

Page No 66:

Question 4:

Find the 19^th term of the following A.P.
7, 13, 19, 25,....

Answer:

The given sequence is 7, 13, 19, 25,....

Here,
First term (a) = 7

Common difference (d) = a₂ – a₁ = 13 – (7) = 6

Now,
$a_{19} = a + (n - 1) d = 7 + (19 - 1) 6 = 7 + (18) 6 = 115$

Hence, the 19^th term of the progression is 115.

Page No 66:

Question 5:

Find the 27^th term of the following A.P.
9, 4, –1, –6, –11,...

Answer:

The given sequence is 9, 4, –1, –6, –11,...

Here,
First term (a) = 9

Common difference (d) = a₂ – a₁ = 4 – (9) = –5

Now,
$a_{27} = a + (n - 1) d = 9 + (27 - 1) (- 5) = 9 + (26) (- 5) = - 121$

Hence, the 27^th term of the progression is –121.

Page No 66:

Question 6:

Find how many three digit natural numbers are divisible by 5.

Answer:

The least positive three digit natural number divisible by 5 is 100.

The sequence divisible by 5 is 100, 105, 110, .... , 995.

Now,
$a_{n} = a + (n - 1) d \Rightarrow 995 = 100 + (n - 1) (5) \Rightarrow 995 = 100 + 5 n - 5 \Rightarrow 995 = 95 + 5 n \Rightarrow 995 - 95 = 5 n \Rightarrow 5 n = 900 \Rightarrow n = 180$

Hence, 180 three digit natural numbers are divisible by 5.

Page No 66:

Question 7:

The 11^th term and the 21^st term of an A.P. are 16 and 29 respectively, then find the 41^th term of that A.P.

Answer:

It is given that,
a₁₁ = 16 and a₂₁ = 29

We know that,
$a_{n} = a + (n - 1) d a_{11} = a + (11 - 1) d \Rightarrow 16 = a + 10 d \Rightarrow a = 16 - 10 d . . . (1) a_{21} = a + (21 - 1) d \Rightarrow 29 = a + 20 d \Rightarrow 29 = 16 - 10 d + 20 d (from (1)) \Rightarrow 29 - 16 = 10 d \Rightarrow d = \frac{13}{10} . . . (2) Putting the value of d in (1), we get a = 16 - 10 (\frac{13}{10}) = 3 ∴ a_{41} = a + (41 - 1) d = 3 + (40) \frac{13}{10} = 55$

Hence, the 41^th term of the A.P. is 55.

Page No 66:

Question 8:

11, 8, 5, 2,.... In this A.P. which term is number –151?

Answer:

The given sequence is 11, 8, 5, 2,.... –151.

Now,
$a_{n} = a + (n - 1) d \Rightarrow - 151 = 11 + (n - 1) (- 3) \Rightarrow - 151 = 11 - 3 n + 3 \Rightarrow - 151 = 14 - 3 n \Rightarrow 3 n = 14 + 151 \Rightarrow 3 n = 165 \Rightarrow n = 55$

Hence, 55^th term is number –151.

Page No 66:

Question 9:

In the natural numbers from 10 to 250, how many are divisible by 4?

Answer:

The least positive natural number from 10 to 250 divisible by 4 is 12.

The sequence divisible by 4 is 12, 16, 20, .... , 248.

Now,
$a_{n} = a + (n - 1) d \Rightarrow 248 = 12 + (n - 1) (4) \Rightarrow 248 = 12 + 4 n - 4 \Rightarrow 248 = 8 + 4 n \Rightarrow 248 - 8 = 4 n \Rightarrow 4 n = 240 \Rightarrow n = 60$

Hence, 60 numbers are divisible by 4.

Page No 66:

Question 10:

In an A.P. 17^th term is 7 more than its 10^th term. Find the common difference.

Answer:

It is given that,
a₁₇ = 7 + a₁₀

We know that,
$a_{n} = a + (n - 1) d a_{17} = 7 + a_{10} \Rightarrow a + (17 - 1) d = 7 + a + (10 - 1) d \Rightarrow a + 16 d = 7 + a + 9 d \Rightarrow 16 d = 7 + 9 d \Rightarrow 16 d - 9 d = 7 \Rightarrow 7 d = 7 \Rightarrow d = 1$

Hence, the common difference is 1.

Page No 72:

Question 1:

First term and common difference of an A.P. are 6 and 3 respectively ; find S₂₇.

Answer:

It is given that,
First term (a) = 6
Common difference (d) = 3

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) ∴ S_{27} = \frac{27}{2} (2 (6) + (27 - 1) (3)) = \frac{27}{2} (12 + 26 (3)) = \frac{27}{2} (12 + 78) = \frac{27}{2} \times 90 = 1215$

Hence, S₂₇ = 1215.

Page No 72:

Question 2:

Find the sum of first 123 even natural numbers.

Answer:

The given sequence is 2, 4, 6, 8 ...
First term (a) = 2
Common difference (d) = 2
Number of terms (n) = 123

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) ∴ S_{123} = \frac{123}{2} (2 (2) + (123 - 1) (2)) = \frac{123}{2} (4 + 122 (2)) = \frac{123}{2} (4 + 244) = \frac{123}{2} \times 248 = 15252$

Hence, the sum of first 123 even natural numbers is 15252.

Page No 72:

Question 3:

Find the sum of all even numbers from 1 to 350.

Answer:

The given sequence is 2, 4, 6, 8 ... 350
First term (a) = 2
Common difference (d) = 2

Now,
$a_{n} = a + (n - 1) d \Rightarrow 350 = 2 + (n - 1) 2 \Rightarrow 350 = 2 + 2 n - 2 \Rightarrow 2 n = 350 \Rightarrow n = 175$

Thus, the number of terms (n) = 175.

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) ∴ S_{175} = \frac{175}{2} (2 (2) + (175 - 1) (2)) = \frac{175}{2} (4 + 174 (2)) = \frac{175}{2} (4 + 348) = \frac{175}{2} \times 352 = 30800$

Hence, the sum of all even numbers from 1 to 350 is 30800.

Page No 72:

Question 4:

In an A.P. 19^th term is 52 and 38^thterm is 128, find sum of first 56 terms.

Answer:

It is given that,
a₁₉ = 52
a₃₈ = 128

Now,
$a_{n} = a + (n - 1) d a_{19} = a + (19 - 1) d \Rightarrow 52 = a + 18 d \Rightarrow a = 52 - 18 d . . . (1) a_{38} = a + (38 - 1) d \Rightarrow 128 = a + 37 d . . . (2) \Rightarrow 128 = 52 - 18 d + 37 d (from (1)) \Rightarrow 128 - 52 = 19 d \Rightarrow 19 d = 76 \Rightarrow d = 4 ∴ a = 52 - 18 (4) = - 20 Thus, a = - 20 and d = 4 .$

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) ∴ S_{56} = \frac{56}{2} (2 (- 20) + (56 - 1) (4)) = 28 (- 40 + 220) = 28 (180) = 5040$

Hence, the sum of first 56 terms is 5040.

Page No 72:

Question 5:

Complete the following activity to find the sum of natural numbers from 1 to 140 which are divisible by 4.

Sum of numbers from 1 to 140, which are divisible by 4 =

Answer:

From 1 to 140, natural numbers divisible by 4 are 4, 8, ... 136.
a = 4
d = 4

Now,
$t_{n} = a + (n - 1) d 136 = 4 + (n - 1) 4 \Rightarrow 136 = 4 + 4 n - 4 \Rightarrow 4 n = 136 \Rightarrow n = 34$

Thus, number of terms (n) = 34.

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) ∴ S_{34} = \frac{34}{2} (2 (4) + (34 - 1) (4)) = 17 (8 + 132) = 17 (140) = 2380$

Hence, the sum of numbers from 1 to 140, which are divisible by 4 = $2380$ .

Page No 72:

Question 6:

Sum of first 55 terms in an A.P. is 3300, find its 28^th term.

Answer:

It is given that,
S₅₅ = 3300

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{55} = \frac{55}{2} (2 a + (55 - 1) d) \Rightarrow 3300 = \frac{55}{2} (2 a + 54 d) \Rightarrow \frac{3300 \times 2}{55} = 2 (a + 27 d) \Rightarrow 120 = 2 (a + (28 - 1) d) \Rightarrow 60 = a + (28 - 1) d \Rightarrow a_{28} = 60$

Hence, its 28^th term is 60.

Page No 73:

Question 7:

In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
( Assume that three consecutive terms in A.P. are a – d , a , a + d .)

Answer:

Assume that three consecutive terms in A.P. are a – d , a , a + d .

It is given that,
Sum of three consecutive terms = 27
Product of three consecutive terms = 504

$(a - d) + a + (a + d) = 27 \Rightarrow 3 a = 27 \Rightarrow a = 9$

$(a - d) \times a \times (a + d) = 504 \Rightarrow (9 - d) \times 9 \times (9 + d) = 504 \Rightarrow 81 - d^{2} = 56 \Rightarrow d^{2} = 81 - 56 \Rightarrow d^{2} = 25 \Rightarrow d = \pm 5$

When d = 5,
The terms are 9, 14, 19.

When d = –5,
The terms are 9, 4, –1.

Page No 73:

Question 8:

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3^rd and 4^th term is 14.

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)

Answer:

Assume that the four consecutive terms in A.P. are a – d, a, a + d, a +2d .

It is given that,
Sum of four consecutive terms = 12
Sum of 3^rd and 4^th term = 14

$(a - d) + a + (a + d) + (a + 2 d) = 12 \Rightarrow 4 a + 2 d = 12 \Rightarrow 2 a + d = 6 \Rightarrow 2 a = 6 - d . . . (1)$

$(a + d) + (a + 2 d) = 14 \Rightarrow 2 a + 3 d = 14 \Rightarrow 6 - d + 3 d = 14 \Rightarrow 2 d = 14 - 6 \Rightarrow 2 d = 8 \Rightarrow d = 4 (from (1)) \Rightarrow 2 a = 6 - d \Rightarrow 2 a = 6 - 4 \Rightarrow 2 a = 2 \Rightarrow a = 1$

Hence, the terms are –3, 1, 5 and 9.

Page No 73:

Question 9:

If the 9^th term of an A.P. is zero then show that the 29^thterm is twice the 19^th term.

Answer:

It is given that,
a₉ = 0

Now,
$a_{n} = a + (n - 1) d a_{9} = a + (9 - 1) d \Rightarrow 0 = a + 8 d \Rightarrow a = - 8 d . . . (1) a_{29} = a + (29 - 1) d = - 8 d + 28 d = 20 d . . . (2) a_{19} = a + (19 - 1) d = - 8 d + 18 d = 10 d . . . (3) From (2) and (3), we get a_{29} = 2 a_{19}$

Hence, the 29^thterm is twice the 19^th term.

Page No 78:

Question 1:

On 1st Jan 2016, Sanika decides to save Rs 10, Rs 11 on second day, Rs 12 on third day. If she decides to save like this, then on 31^st Dec 2016 what would be her total saving ?

Answer:

On 1^st Jan 2016, Sanika saved Rs 10.
Next day Rs 11.
Third day Rs 12.

Her total savings = Rs (10 + 11 + 12 + ... + till 31^st Dec)

Here,
a = 10
d = 1
n = 366 (since 2016 is a leap year)

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) \Rightarrow S_{366} = \frac{366}{2} (2 (10) + (366 - 1) 1) = 183 (20 + 365) = 70455$

Hence, her total savings would be Rs 70455.

Page No 78:

Question 2:

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly instalments. Each instalment being less than the preceding one by Rs 40. Find the amount of the first and last instalment.

Answer:

Money he borrows = Rs 8000
Total interest = Rs 1360

Total money he will pay after 12 months = Rs (8000 + 1360) = Rs 9360

It is given that,
Each instalment being less than the preceding one by Rs 40.

d = −40
n = 12
S₁₂ = 9360

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{12} = \frac{12}{2} (2 a + (12 - 1) d) \Rightarrow 9360 = \frac{12}{2} (2 (a) + (12 - 1) (- 40)) \Rightarrow 9360 = 6 (2 a - 440) \Rightarrow 2 a - 440 = \frac{9360}{6} \Rightarrow 2 a - 440 = 1560 \Rightarrow 2 a = 1560 + 440 \Rightarrow 2 a = 2000 \Rightarrow a = 1000$

Also,
$a_{12} = a + (12 - 1) d = 1000 + 11 (- 40) = 1000 - 440 = 560$

Hence, the first and last instalment are Rs 1000 and Rs 560, respectively.

Page No 78:

Question 3:

Sachin invested in a national saving certificate scheme. In the first year he invested Rs 5000 , in the second year Rs 7000, in the third year Rs 9000 and so on. Find the total amount that he invested in 12 years.

Answer:

Sachin invested in first year = Rs 5000
Second year investment = Rs 7000
Third year investment = Rs 9000

Total investment = Rs (5000 + 7000 + 9000 + ..... + in 12 years)

Here,
a = 5000
d = 2000
n = 12

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{12} = \frac{12}{2} (2 a + (12 - 1) d) = \frac{12}{2} (2 (5000) + (12 - 1) (2000)) = 6 (10000 + 22000) = 6 (32000) = 192000$

Hence, the total amount that he invested in 12 years is Rs 192000.

Page No 78:

Question 4:

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15^th row and also find how many total seats are there in the auditorium ?

Answer:

Total number of rows = 27
Number of seats in first row = 20
Number of seats in second row = 22
Number of seats in third row = 24

Total number of seats = 20 + 22 + 24 + ..... + upto 27 rows

Here,
a = 20
d = 2
n = 27

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{27} = \frac{27}{2} (2 a + (27 - 1) d) = \frac{27}{2} (2 (20) + 26 (2)) = \frac{27}{2} (40 + 52) = \frac{27}{2} (92) = 27 \times 46 = 1242$

Hence, there are 1242 seats in the auditorium.

Also,
$a_{15} = a + (15 - 1) d = 20 + 14 (2) = 20 + 28 = 48$

Hence, the number of seats in the 15^th row is 48.

Page No 78:

Question 5:

Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was –30° celsius then find the temperature on the other five days.

Answer:

The readings of the temperature were in A.P.
Let, first term = a
common difference = d

Temperature of Monday = a
Temperature of Tuesday = a + d
Temperature of Wednesday = a + 2d
Temperature of Thursday = a + 3d
Temperature of Friday = a + 4d
Temperature of Saturday = a + 5d

The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday.
$∴ a + (a + 5 d) = 5 + (a + d + a + 5 d) \Rightarrow 2 a + 5 d = 5 + 2 a + 6 d \Rightarrow 6 d - 5 d = - 5 \Rightarrow d = - 5$

Also, the temperature of Wednesday was –30° celsius.
$∴ a + 2 d = - 30 \Rightarrow a + 2 (- 5) = - 30 \Rightarrow a - 10 = - 30 \Rightarrow a = - 30 + 10 \Rightarrow a = - 20$

Hence, the temperature on the other five days were:
Monday = –20° celsius
Tuesday = –25° celsius
Thursday = –35° celsius
Friday = –40° celsius
Saturday = –45° celsius

Page No 78:

Question 6:

On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Answer:

Total number of rows = 25
Number of trees in first row = 1
Number of trees in second row = 2
Number of trees in third row = 3

Total number of trees = 1 + 2 + 3 + ..... + upto 25 rows

Here,
a = 1
d = 1
n = 25

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{25} = \frac{25}{2} (2 a + (25 - 1) d) = \frac{25}{2} (2 (1) + 24 (1)) = \frac{25}{2} (2 + 24) = \frac{25}{2} (26) = 25 \times 13 = 325$

Hence, the total number of trees are 325.

Page No 78:

Question 1:

Choose the correct alternative answer for each of the following sub questions.
(1) The sequence –10, –6, –2, 2,...

(A) is an A.P., Reason d= –16	(B) is an A.P., Reason d= 4
(C) is an A.P., Reason d= –4	(D) is not an A.P.

(2) First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.

(A) –2, 0, 2, 4	(B) –2, 4, –8, 16
(C) –2, –4, –6, –8	(D) –2, –4, –8, –16

(3) What is the sum of the first 30 natural numbers ?

(A) 464

(B) 465

(D) 461

(4) For an given A.P. t₇ = 4, d = –4, n = 101, then a = ....

(A) 6

(B) 7

(D) 28

(5) For an given A.P. a = 3.5, d = 0, n = 101, then t_n = ....

(A) 0

(B) 3.5

(D) 14.5

(6) In an A.P. first two terms are –3, 4 then 21^st term is ...

(A) –143

(B) 143

(D) 17

(7) If for any A.P. d = 5 then t₁₈ – t₁₃ = ....

(A) 5

(B) 20

(D) 30

(8) Sum of first five multuiples of 3 is...

(A) 45

(B) 55

(D) 75

(9) 15, 10, 5,... In this A.P sum of first 10 terms is...

(A) –75

(B) –125

(D) 125

(10) In an A.P. 1^st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....

(A) 42

(B) 38

(D) 19

Answer:

(1) The given sequence is –10, –6, –2, 2,...

Here,
First term (a) = a₁ = –10
Second term = a₂ = –6
Third term = a₃ = –2

Common difference (d) = a₂ – a₁ = –6 – (–10) = 4
= a₃ – a₂ = –2 – (–6) = 4

Since, a₂ – a₁ = a₃ – a₂

Thus, the given sequence is an A.P.

Hence, the correct option is (B).

(2) It is given that,
First term (a) = –2
Common difference (d) = –2

Second term = a + d = –2 + (–2) = –4
Third term = a + 2d = –2 + 2(–2) = –6
Fourth term = a + 3d = –2 + 3(–2) = –8

Thus, first four terms of the A.P. are –2, –4, –6, –8

Hence, the correct option is (C).

(3) The given series is 1 + 2 + 3 + ... + 30

Here,
a = 1
d = 1
n = 30

$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{30} = \frac{30}{2} (2 a + (30 - 1) d) = \frac{30}{2} (2 (1) + 29 (1)) = 15 (2 + 29) = 15 \times 31 = 465$

Hence, the correct option is (B).

(4) It is given that,
t₇ = 4
d = –4
n = 101

Now,
$t_{n} = a + (n - 1) d t_{7} = a + (7 - 1) d \Rightarrow 4 = a + 6 (- 4) \Rightarrow 4 = a - 24 \Rightarrow a = 4 + 24 \Rightarrow a = 28$

Hence, the correct option is (D).

(5) It is given that,
a = 3.5
d = 0
n = 101

$t_{n} = a + (n - 1) d = 3.5 + (n - 1) (0) = 3.5$

Hence, the correct option is (B).

(6) It is given that,
a = –3
a₂ = 4

We know that,
$a_{2} = a + (2 - 1) d \Rightarrow 4 = - 3 + d \Rightarrow d = 7$

Now,
$a_{21} = a + (21 - 1) d = - 3 + 20 (7) = - 3 + 140 = 137$

Hence, the correct option is (C).

(7) It is given that,
d = 5

Now,
$t_{n} = a + (n - 1) d t_{18} - t_{13} = (a + (18 - 1) d) - (a + (13 - 1) d) = (a + 17 d) - (a + 12 d) = 5 d = 5 (5) = 25$

Hence, the correct option is (C).

(8) The given sequence is 3, 6, 9,...
Here,
a = 3
d = 3
n = 5

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{5} = \frac{5}{2} (2 a + (5 - 1) d) = \frac{5}{2} (2 (3) + 4 (3)) = \frac{5}{2} (6 + 12) = \frac{5}{2} (18) = 5 (9) = 45$

Hence, the correct option is (A).

(9) The given sequence is 15, 10, 5,...
Here,
a = 15
d = –5

We know that,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{10} = \frac{10}{2} (2 a + (10 - 1) d) = 5 (2 (15) + 9 (- 5)) = 5 (30 - 45) = 5 (- 15) = - 75$

Hence, the correct option is (A).

(10) It is given that,
First term (a) = 1
Last term (t_n) = 20
Sum of terms (S_n) = 399

We know that,
$t_{n} = a + (n - 1) d S_{n} = \frac{n}{2} (2 a + (n - 1) d) \Rightarrow S_{n} = \frac{n}{2} (a + (a + (n - 1) d)) \Rightarrow S_{n} = \frac{n}{2} (a + t_{n}) \Rightarrow 399 = \frac{n}{2} (1 + 20) \Rightarrow 399 = \frac{21 n}{2} \Rightarrow 21 n = 399 \times 2 \Rightarrow n = \frac{798}{21} \Rightarrow n = 38$

Hence, the correct option is (B).

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Question 2:

Find the fourth term from the end in an A.P. –11, –8, –5,...., 49.

Answer:

The given sequence is –11, –8, –5,...., 49.

Here,
First term (a) = –11
Common difference (d) = 3
n = 4
Last term (l) = 49

We know that,
$n^{th} term from the end = l - (n - 1) d 4^{th} term from the end = l - (4 - 1) d = 49 - 3 (3) = 49 - 9 = 40$

Hence, the fourth term from the end is 40.

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Question 3:

In an A.P. the 10^th term is 46 sum of the 5^th and 7^th term is 52. Find the A.P.

Answer:

It is given that,
t₁₀ = 46
t₅ + t₇ = 52

Now,
$t_{n} = a + (n - 1) d t_{10} = a + (10 - 1) d \Rightarrow 46 = a + 9 d \Rightarrow a = 46 - 9 d . . . (1) t_{5} + t_{7} = 52 \Rightarrow (a + (5 - 1) d) + (a + (7 - 1) d) = 52 \Rightarrow a + 4 d + a + 6 d = 52 \Rightarrow 2 a + 10 d = 52 \Rightarrow 2 (46 - 9 d) + 10 d = 52 (from (1)) \Rightarrow 92 - 18 d + 10 d = 52 \Rightarrow 92 - 8 d = 52 \Rightarrow 8 d = 92 - 52 \Rightarrow 8 d = 40 \Rightarrow d = 5 \Rightarrow a = 46 - 9 (5) (from (1)) \Rightarrow a = 46 - 45 \Rightarrow a = 1$

Hence, the given A.P. is 1, 6, 11, 16, ....

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Question 4:

The A.P. in which 4^th term is –15 and 9^th term is –30. Find the sum of the first 10 numbers.

Answer:

It is given that,
t₄ = –15
t₉ = –30

Now,
$t_{n} = a + (n - 1) d t_{4} = a + (4 - 1) d \Rightarrow - 15 = a + 3 d \Rightarrow a = - 15 - 3 d . . . (1) t_{9} = a + (9 - 1) d \Rightarrow - 30 = a + 8 d \Rightarrow a + 8 d = - 30 \Rightarrow - 15 - 3 d + 8 d = - 30 (from (1)) \Rightarrow - 15 + 5 d = - 30 \Rightarrow 5 d = - 30 + 15 \Rightarrow 5 d = - 15 \Rightarrow d = - 3 \Rightarrow a = - 15 - 3 (- 3) (from (1)) \Rightarrow a = - 15 + 9 \Rightarrow a = - 6$

Hence, the given A.P. is –6, –9, –12, ....

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) S_{10} = \frac{10}{2} (2 a + (10 - 1) d) = 5 (2 (- 6) + 9 (- 3)) = 5 (- 12 - 27) = 5 (- 39) = - 195$

Hence, the sum of the first 10 numbers is –195.

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Question 5:

Two A.P.’ s are given 9, 7, 5, . . . and 24, 21, 18, . . . . If n^th term of both the progressions are equal then find the value of n and n th term.

Answer:

The given sequence is
9, 7, 5, . . .

Here,
a = 9
d = –2

We know that,
$a_{n} = a + (n - 1) d = 9 + (n - 1) (- 2) = 9 - 2 n + 2 = 11 - 2 n . . . (1)$

Another given sequence is
24, 21, 18, . . . .

Here,
a = 24
d = –3

We know that,
${a^{'}}_{n} = a + (n - 1) d = 24 + (n - 1) (- 3) = 24 - 3 n + 3 = 27 - 3 n . . . (2)$

It is given that,
n^th term of both the progressions are equal.

From (1) and (2), we get
$11 - 2 n = 27 - 3 n \Rightarrow 3 n - 2 n = 27 - 11 \Rightarrow n = 16$

Hence, n = 16.

Now,
$a_{16} = 11 - 2 (16) (from (1)) = 11 - 32 = - 21$

Hence, the value of n and n^th term is 16 and –21, respectively.

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Question 6:

If sum of 3^rd and 8^th terms of an A.P. is 7 and sum of 7^th and 14^th terms is –3 then find the 10 th term .

Answer:

It is given that,
a₃ + a₈ = 7
a₇ + a₁₄ = –3

Now,
$a_{n} = a + (n - 1) d a_{3} + a_{8} = 7 \Rightarrow (a + (3 - 1) d) + (a + (8 - 1) d) = 7 \Rightarrow a + 2 d + a + 7 d = 7 \Rightarrow 2 a + 9 d = 7 \Rightarrow 2 a = 7 - 9 d . . . (1) a_{7} + a_{14} = - 3 \Rightarrow (a + (7 - 1) d) + (a + (14 - 1) d) = - 3 \Rightarrow a + 6 d + a + 13 d = - 3 \Rightarrow 2 a + 19 d = - 3 \Rightarrow 7 - 9 d + 19 d = - 3 (from (1)) \Rightarrow 7 + 10 d = - 3 \Rightarrow 10 d = - 3 - 7 \Rightarrow 10 d = - 10 \Rightarrow d = - 1 \Rightarrow 2 a = 7 - 9 (- 1) (from (1)) \Rightarrow 2 a = 7 + 9 \Rightarrow 2 a = 16 \Rightarrow a = 8$

Now,
$a_{10} = a + (10 - 1) d = 8 + 9 (- 1) = 8 - 9 = - 1$

Hence, the 10^th term is –1.

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Question 7:

In an A.P. the first term is –5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there ? What is the common difference ?

Answer:

It is given that,
a = –2
l = 45
S_n= 120

Now,
$S_{n} = \frac{n}{2} (a + l) \Rightarrow 120 = \frac{n}{2} (- 5 + 45) \Rightarrow 120 = \frac{n}{2} (40) \Rightarrow 120 \times 2 = n (40) \Rightarrow 240 = n (40) \Rightarrow n = \frac{240}{40} \Rightarrow n = 6$

Hence, there are 6 terms.

Also,
$l = a + (6 - 1) d \Rightarrow 45 = - 5 + 5 d \Rightarrow 45 + 5 = 5 d \Rightarrow 5 d = 50 \Rightarrow d = 10$

Hence, the common difference is 10.

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Question 8:

Sum of 1 to n natural numbers is 36, then find the value of n .

Answer:

It is given that,
a = 1
d = 1
S_n= 36

Now,
$S_{n} = \frac{n}{2} (2 a + (n - 1) d) \Rightarrow 36 = \frac{n}{2} (2 (1) + (n - 1) (1)) \Rightarrow 36 \times 2 = n (2 + n - 1) \Rightarrow 72 = n (n + 1) \Rightarrow n^{2} + n - 72 = 0 \Rightarrow n^{2} + 9 n - 8 n - 72 = 0 \Rightarrow n (n + 9) - 8 (n + 9) = 0 \Rightarrow (n + 9) (n - 8) = 0 \Rightarrow n = - 9 or n = 8 \Rightarrow n = 8 (∵ n \neq - 9)$

Hence, the value of n is 8.

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Question 9:

Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.

Answer:

Let the three numbers be a – d, a and a + d.

According to the question,
$(a - d) + a + (a + d) = 207 \Rightarrow 3 a = 207 \Rightarrow a = 69$

Also,
$(a - d) a = 4623 \Rightarrow (69 - d) (69) = 4623 \Rightarrow 69 - d = \frac{4623}{69} \Rightarrow 69 - d = 67 \Rightarrow 69 - 67 = d \Rightarrow d = 2$

Hence, the three numbers are 67, 69 and 71.

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Question 10:

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Answer:

Let the first term be a and the common difference be d.

Since, the A.P. contains 37 terms, the middle term is ${(\frac{37 + 1}{2})}^{th} = 19^{th}$ term.

According to the question,
$a_{18} + a_{19} + a_{20} = 225 \Rightarrow (a + (18 - 1) d) + (a + (19 - 1) d) + (a + (20 - 1) d) = 225 \Rightarrow 3 a + 17 d + 18 d + 19 d = 225 \Rightarrow 3 a + 54 d = 225 \Rightarrow 3 (a + 18 d) = 225 \Rightarrow a + 18 d = \frac{225}{3} \Rightarrow a + 18 d = 75 \Rightarrow a = 75 - 18 d . . . (1)$

Also,
$a_{35} + a_{36} + a_{37} = 429 \Rightarrow (a + (35 - 1) d) + (a + (36 - 1) d) + (a + (37 - 1) d) = 429 \Rightarrow 3 a + 34 d + 35 d + 36 d = 429 \Rightarrow 3 a + 105 d = 429 \Rightarrow 3 (a + 35 d) = 429 \Rightarrow a + 35 d = \frac{429}{3} \Rightarrow 75 - 18 d + 35 d = 143 (from (1)) \Rightarrow 17 d = 143 - 75 \Rightarrow 17 d = 68 \Rightarrow d = 4 \Rightarrow a = 75 - 18 (4) (from (1)) \Rightarrow a = 75 - 72 \Rightarrow a = 3$

Hence, the resulting A.P is 3, 7, 11, ....

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Question 11:

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$ .

Answer:

It is given that,
First term = a
a₂ = b
l = c

Now,
$a_{n} = a + (n - 1) d a_{2} = a + (2 - 1) d \Rightarrow b = a + d \Rightarrow d = b - a . . . (1) l = a + (n - 1) d \Rightarrow c = a + (n - 1) (b - a) (from (1)) \Rightarrow c - a = (n - 1) (b - a) \Rightarrow (n - 1) = \frac{c - a}{b - a} \Rightarrow n = \frac{c - a}{b - a} + 1 \Rightarrow n = \frac{c - a + b - a}{b - a} \Rightarrow n = \frac{c + b - 2 a}{b - a} . . . (2)$

Also,
$S_{n} = \frac{n}{2} (a + l) = \frac{(\frac{c + b - 2 a}{b - a})}{2} (a + c) (from (2)) = \frac{(c + b - 2 a) (a + c)}{2 (b - a)}$

Hence, the sum of all terms is $\frac{(a + c) (b + c - 2 a)}{2 (b - a)}$ .

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Question 12:

If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p + q) terms is zero. (p ≠ q)

Answer:

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

According to the question,
$S_{p} = S_{q} \Rightarrow \frac{p}{2} (2 a + (p - 1) d) = \frac{q}{2} (2 a + (q - 1) d) \Rightarrow p (2 a + (p - 1) d) = q (2 a + (q - 1) d) \Rightarrow 2 a p + (p - 1) p d = 2 a q + (q - 1) q d \Rightarrow 2 a p + p^{2} d - p d = 2 a q + q^{2} d - q d \Rightarrow 2 a p - 2 a q = q^{2} d - q d - p^{2} d + p d \Rightarrow 2 a (p - q) = d (q^{2} - p^{2}) + d (p - q) \Rightarrow 2 a (p - q) = d (q + p) (q - p) + d (p - q) \Rightarrow 2 a (p - q) = d [(q + p) (q - p) + (p - q)] \Rightarrow 2 a (p - q) = d [- (q + p) (p - q) + (p - q)] \Rightarrow 2 a (p - q) = d (p - q) [1 - q - p] \Rightarrow 2 a = d (1 - q - p) (∵ p \neq q) \Rightarrow 2 a = d (1 - q - p) . . . (1)$

Now,
$S_{p + q} = (\frac{p + q}{2}) (2 a + (p + q - 1) d) = (\frac{p + q}{2}) ((1 - q - p) d + (p + q - 1) d) (from (1)) = (\frac{p + q}{2}) d (1 - q - p + p + q - 1) = 0$

Hence, the sum of its first (p + q) terms is zero.

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Question 13:

If m times the m^th term of an A.P. is eqaul to n times n^th term then show that the (m + n)^thterm of the A.P. is zero.

Answer:

We know,
$a_{n} = a + (n - 1) d$

According to the question,

$m (a_{m}) = n (a_{n}) \Rightarrow m (a + (m - 1) d) = n (a + (n - 1) d) \Rightarrow a m + (m - 1) m d = a n + (n - 1) n d \Rightarrow a m + m^{2} d - m d = a n + n^{2} d - n d \Rightarrow a m - a n = n^{2} d - n d - m^{2} d + m d \Rightarrow a (m - n) = d (n^{2} - m^{2}) + d (m - n) \Rightarrow a (m - n) = d (m + n) (n - m) + d (m - n) \Rightarrow a (m - n) = d [(m + n) (n - m) + (m - n)] \Rightarrow a (m - n) = d [- (m + n) (m - n) + (m - n)] \Rightarrow a (m - n) = d (m - n) [1 - m - n] \Rightarrow a = d (1 - m - n) (∵ m \neq n) \Rightarrow a = d (1 - m - n) . . . (1)$

Now,
$a_{m + n} = (a + (m + n - 1) d) = ((1 - m - n) d + (m + n - 1) d) (from (1)) = d (1 - m - n + m + n - 1) = 0$

Hence, the (m + n)^thterm of the A.P. is zero.

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Question 14:

Rs 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.

Answer:

It is given that,
Principal (P) = Rs 1000
Rate (R) = 10%
Simple interest (S.I.) = $\frac{P \times R \times T}{100}$

Simple interest after an year = $\frac{1000 \times 10 \times 1}{100} = Rs 100$

Simple interest after 2 years = $\frac{1000 \times 10 \times 2}{100} = Rs 200$

Simple interest after 3 years = $\frac{1000 \times 10 \times 3}{100} = Rs 300$

Hence, the total interest amount is in A.P. i.e. 100, 200, 300,....

Here,
a = 100
d = 100

Now,
$a_{n} = a + (n - 1) d a_{20} = a + (20 - 1) d = 100 + 19 (100) = 100 + 1900 = 2000$

Hence, the interest amount after 20 years is Rs 2000.

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