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#### Question 1:

Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.

(i) 9, 15, 21, 27, ...
(ii) 11, 6, 1, −4, ...
(iii) −1, $\frac{-5}{6}$, $\frac{-2}{3}$, $\frac{-1}{2}$, ...
(iv)
(v)

(i) The given progression 9, 15, 21, 27, ... .

Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an AP.

First term = 9

Common difference = 6

Next term of the AP = 27 + 6 = 33

(ii) The given progression 11, 6, 1, −4, ... .

Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)

Thus, each term differs from its preceding term by −5. So, the given progression is an AP.

First term = 11

Common difference = −5

Next term of the AP = −4 + (−5) = −9

(iii) The given progression −1, $\frac{-5}{6}$, $\frac{-2}{3}$, $\frac{-1}{2}$, ...

Clearly, $\frac{-5}{6}-\left(-1\right)=\frac{-2}{3}-\left(\frac{-5}{6}\right)=\frac{-1}{2}-\left(\frac{-2}{3}\right)=\frac{1}{6}$ (Constant)

Thus, each term differs from its preceding term by $\frac{1}{6}$. So, the given progression is an AP.

First term = −1

Common difference = $\frac{1}{6}$

Next term of the AP = $\frac{-1}{2}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}$

(iv) The given progression

This sequence can be re-written as

Clearly, $2\sqrt{2}-\sqrt{2}=3\sqrt{2}-2\sqrt{2}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{2}$. So, the given progression is an AP.

First term = $\sqrt{2}$

Common difference = $\sqrt{2}$

Next term of the AP = $4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}$

(v) The given progression

This sequence can be re-written as

Clearly, $3\sqrt{5}-2\sqrt{5}=4\sqrt{5}-3\sqrt{5}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{5}$. So, the given progression is an AP.

First term = $2\sqrt{5}=\sqrt{20}$

Common difference = $\sqrt{5}$

Next term of the AP = $5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$

#### Question 2:

Find:

(i) the 20th term of the AP 9, 13, 17, 21, ... .
(ii) the 35th term of the AP 20, 17, 14, 11, ... .
(iii) the 18th term of the AP , ... .
(iv) the 9th term of the AP $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .
(v) the 15th term of the AP −40, −15, 10, 35, ... .

(i)  The given AP is 9, 13, 17, 21, ... .

First term, a = 9

Common difference, d = 13 − 9 = 4

nth term of the AP, an = a + (− 1)d = 9 + (− 1) × 4

∴ 20th term of the AP, a20 = 9 + (20 − 1) × 4 = 9 + 76 = 85

(ii)  The given AP is 20, 17, 14, 11, ... .

First term, a = 20

Common difference, d = 17 − 20 = −3

nth term of the AP, an = a + (− 1)d = 20 + (− 1) × (−3)

∴ 35th term of the AP, a35 = 20 + (35 − 1) × (−3) = 20 − 102 = −82

(iii)  The given AP is , ... .

This can be re-written as , ... .

First term, a = $\sqrt{2}$

Common difference, d = $3\sqrt{2}-\sqrt{2}=2\sqrt{2}$

nth term of the AP, an = a + (− 1)d = $\sqrt{2}+\left(n-1\right)×2\sqrt{2}$

∴ 18th term of the AP, a18 = $\sqrt{2}+\left(18-1\right)×2\sqrt{2}=\sqrt{2}+34\sqrt{2}=35\sqrt{2}=\sqrt{2450}$

(iv)  The given AP is $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .

First term, a = $\frac{3}{4}$

Common difference, d = $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$

nth term of the AP, an = a + (− 1)d = $\frac{3}{4}+\left(n-1\right)×\left(\frac{1}{2}\right)$

∴ 9th term of the AP, a9 = $\frac{3}{4}+\left(9-1\right)×\frac{1}{2}=\frac{3}{4}+4=\frac{19}{4}$

(v)  The given AP is −40, −15, 10, 35, ... .

First term, a = −40

Common difference, d = −15 − (−40) = 25

nth term of the AP, an = a + (− 1)d = −40 + (− 1) × 25

∴ 15th term of the AP, a15 = −40 + (15 − 1) × 25 = −40 + 350 = 310

#### Question 3:

Find the 37th term of the AP

The given AP is
First term, a = 6 and common difference, d

#### Question 4:

Find the 25th term of the AP

The given AP is
First term = 5
Common difference

#### Question 6:

If the nth term of a progression is (4n − 10), show that it is an AP. Find its
(i) first term, (ii) common difference, and (iii) 16th term.

Tn = (4n - 10)     [Given]
​T1 = (4 ⨯ 1 - 10) = -6
T2 = (4 ⨯ 2 - 10) = -2
T3 = (4 ⨯ 3 - 10) = 2
T4 = (4 ⨯ 4 - 10) = 6

Clearly, [ -2 - (-6)] = [2 - (-2)] = [6 - 2] = 4               (Constant)
So, the terms -6, -2, 2, 6,... forms an AP.

Thus, we have;
(i) First term = -6
(ii) Common difference = 4
(iii) T16 = a + (n -1)d  = a + 15d = ​- 6 + 15 ⨯ 4 = 54

#### Question 7:

How many terms are there in the AP 6, 10, 14, 18,...,174?

In the given AP, a  = 6 and d = (10 - 6) = 4
Suppose that there are n terms in the given AP.
Then Tn = 174
a + (n - 1)d = 174
⇒ 6 + (n - 1) ⨯ 4 = 174
⇒ 2 + 4n = 174
⇒ 4n = 172
n = 43
Hence, there are 43 terms in the given AP.

#### Question 8:

How many terms are there in the AP 41, 38, 35,...., 8 ?

In the given AP, a  = 41 and d = (38 - 41) = -3
Suppose that there are n terms in the given AP.
Then Tn = 8
⇒ a + (n - 1) d = 8
⇒ 41 + (n - 1) ⨯ (-3) = 8
⇒ 44 - 3n = 8
⇒ 3n = 36
n = 12
Hence, there are 12 terms in the given AP.

#### Question 9:

How many terms are there in the AP 18, $15\frac{1}{2}$, 13, ..., −47?

The given AP is 18, $15\frac{1}{2}$, 13, ..., −47.

First term, a = 18

Common difference, d = $15\frac{1}{2}-18=\frac{31}{2}-18=\frac{31-36}{2}=-\frac{5}{2}$

Suppose there are n terms in the given AP. Then,

$⇒n=26+1=27$

Hence, there are 27 terms in the given AP.

#### Question 10:

Which term of the AP 3, 8, 13, 18,... is 88?

In the given AP, first term, a = 3 and common difference, d = (8 - 3) = 5.

Let's its nth term be 88.
Then Tn = 88
⇒ a + (n - 1)d = 88
⇒ 3 + (n - 1) ⨯ 5 = 88
⇒ 5n - 2 = 88
⇒ 5n = 90
n = 18
Hence, the 18th term of the given AP is 88.

#### Question 11:

Which term of the AP 72, 68, 64, 60,... is 0?

In the given AP, first term, a = 72 and common difference, d = (68 - 72) = -4.
Let its nth term be 0.
Then, Tn = 0
⇒ a + (n - 1)d = 0
⇒ 72 + (n - 1) ⨯ (-4) = 0
⇒ 76 - 4n = 0
⇒ 4n = 76
n = 19
Hence, the 19th term of the given AP is 0.

#### Question 12:

Which term of the AP $\frac{5}{6},$ 1, $1\frac{1}{6},1\frac{1}{3},$.... is 3?

In the given AP, first term = $\frac{5}{6}$ and common difference, d .
Let its nth term be 3.

Hence, the 14th term of the given AP is 3.

#### Question 13:

Which term of the AP 21, 18, 15, ... is −81?

The given AP is 21, 18, 15, ... .

First term, a = 21

Common difference, d = 18 − 21 = −3

Suppose nth term of the given AP is −81. Then,

$⇒n=34+1=35$

Hence, the 35th term of the given AP is −81.

#### Question 14:

Which term of the AP 3, 8, 13, 18, ... will be 55 more than its 20th term?

Here, a = 3 and d = (8 - 3) = 5
The 20th term is given by
T20 = a + (20 - 1)d = a + 19d = 3 + 19 ⨯ 5 = 98
∴ Required term = (98 + 55) = 153
Let this be the nth term.
Then Tn = 153
⇒ 3 + (n - 1) ⨯ 5 = 153
⇒ 5n = 155
n = 31
Hence, the 31st term will be 55 more than its 20th term.

#### Question 15:

Which term of the AP 5, 15, 25, ...will be 130 more than its 31st term?

Here, a = 5 and d = (15 - 5) = 10
The 31st term is given by
T31 = a + (31 - 1)d = a + 30d = 5 + 30 ⨯ 10 = 305
∴ Required term = (305 + 130) = 435
Let this be be the nth term.
Then Tn = 435
​⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
n = 44
Hence, the 44th term will be 130 more than its 31st term.

#### Question 16:

If the 10th term of an AP is 52 and the 17th term is 20 more than the 13th term, find the AP.

In the given AP, let the first term be a and the common difference be d.
Then, Tna + (n - 1)d ​
Now, we have:
T10 = a + (10 - 1)d
⇒ a + 9d  = 52       ...(1)
T13a + (13 - 1)d = a + 12d               ...(2)
T17 = a + (17 - 1)d = a + 16d                 ...(3)

But, it is given that T17 = 20 + T13
i.e., a + 16d  = 20 + a + 12d
⇒ 4d = 20
d = 5
On substituting d = 5 in (1), we get:
a + 9 ⨯ 5 = 52
⇒​ a = 7

Thus, a = 7 and d = 5

∴ The terms of the AP are 7, 12, 17, 22,...

#### Question 17:

Find the middle term of the AP 6, 13, 20, ..., 216.                         [CBSE 2015]

The given AP is 6, 13, 20, ..., 216.

First term, a = 6

Common difference, d = 13 − 6 = 7

Suppose there are n terms in the given AP. Then,

$⇒n=30+1=31$

Thus, the given AP contains 31 terms.

∴ Middle term of the given AP

= $\left(\frac{31+1}{2}\right)$th term

= 16th term

= 6 + (16 − 1) × 7

= 6 + 105

= 111

Hence, the middle term of the given AP is 111.

#### Question 18:

Find the middle term of the AP 10, 7, 4, ..., (−62).                         [CBSE 2009C]

The given AP is 10, 7, 4, ..., −62.

First term, a = 10

Common difference, d = 7 − 10 = −3

Suppose there are n terms in the given AP. Then,

$⇒n=24+1=25$

Thus, the given AP contains 25 terms.

∴ Middle term of the given AP

= $\left(\frac{25+1}{2}\right)$th term

= 13th term

= 10 + (13 − 1) × (−3)

= 10 − 36

= −26

Hence, the middle term of the given AP is −26.

#### Question 19:

Find the sum of two middle most terms of the AP $-\frac{4}{3},-1,\frac{-2}{3},...,4\frac{1}{3}$.                     [CBSE 2013C]

The given AP is $-\frac{4}{3},-1,\frac{-2}{3},...,4\frac{1}{3}$.

First term, a = $-\frac{4}{3}$

Common difference, d = $-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{1}{3}$

Suppose there are n terms in the given AP. Then,

$⇒n=17+1=18$

Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.

The middle terms of the given AP are $\left(\frac{18}{2}\right)$th term and $\left(\frac{18}{2}+1\right)$th term i.e. 9th term and 10th term.

∴ Sum of the middle most terms of the given AP

= 9th term + 10th term

$=\left[-\frac{4}{3}+\left(9-1\right)×\frac{1}{3}\right]+\left[-\frac{4}{3}+\left(10-1\right)×\frac{1}{3}\right]\phantom{\rule{0ex}{0ex}}=-\frac{4}{3}+\frac{8}{3}-\frac{4}{3}+3\phantom{\rule{0ex}{0ex}}=3$

Hence, the sum of the middle most terms of the given AP is 3.

#### Question 20:

Find the 8th term from the end of the AP 7, 10, 13, ..., 184.

Here, a = 7 and d = (10 - 7) = 3, l = 184 and n = 8th from the end.
Now, nth term from the end = [l - (n -1)d]
8th term from the end = [184 - (8 - 1) ⨯ 3]
= [184 - (7 ⨯ 3)] = (184 - 21) = 163
Hence, the 8th term from the end is 163.

#### Question 21:

Find the 6th term from the end of the AP 17, 14, 11, ..., (−40).

Here, a = 17 and d = (14 - 17) = -3, l = (-40) and n = 6
Now, nth term from the end = [l - (n - 1)d]
6th term from the end = [(-40) - (6 - 1) ⨯ (-3)]
= [-40 + (5 ⨯ 3)] = (-40 + 15) = -25
Hence, the 6th term from the end is -25.

#### Question 22:

Is 184 a term of the AP 3, 7, 11, 15, ...?

The given AP is 3, 7, 11, 15, ... .

Here, a = 3 and d = 7 − 3 = 4

Let the nth term of the given AP be 184. Then,

$⇒n=\frac{185}{4}=46\frac{1}{4}$

But, the number of terms cannot be a fraction.

Hence, 184 is not a term of the given AP.

#### Question 23:

Is −150 a term of the AP 11, 8, 5, 2, ...?

The given AP is 11, 8, 5, 2, ... .

Here, a = 11 and d = 8 − 11 = −3

Let the nth term of the given AP be −150. Then,

$⇒n=\frac{164}{3}=54\frac{2}{3}$

But, the number of terms cannot be a fraction.

Hence, −150 is not a term of the given AP.

#### Question 24:

Which term of the AP 121, 117, 113, ... is its first negative term?

The given AP is 121, 117, 113, ... .

Here, a = 121 and d = 117 − 121 = −4

Let the nth term of the given AP be the first negative term. Then,

Hence, the 32nd term is the first negative term of the given AP.

#### Question 25:

Which term of the AP 20, $19\frac{1}{4}$$18\frac{1}{2}$, $17\frac{3}{4}$, ... is its first negative term?

The given AP is 20, $19\frac{1}{4}$$18\frac{1}{2}$, $17\frac{3}{4}$, ... .

Here, a = 20 and d = $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$

Let the nth term of the given AP be the first negative term. Then,

Hence, the 28th term is the first negative term of the given AP.

#### Question 26:

The 7th term of an AP is −4 and its 13th term is −6. Find the AP.

We have:
T7 = a + (n - 1)d
⇒ a + 6d = -4           ...(1)

​T13 = a + (n - 1)d
⇒ a + 12d = -16          ...(2)
On solving (1) and (2), we get:
a = 8 and d = -2

Thus, first term = 8 and common difference = -2

∴ The terms of the AP are 8, 6, 4, 2,...

#### Question 27:

The 4th term of an AP is zero. Prove that its 25 term is triple is 11th term.

In the given AP, let the first term be a and the common difference be d.
Then, Tn = a + (n - 1)d ​
Now, T4 = a + (4 - 1)d
⇒ a + 3d  = 0        ...(1)

⇒ a = -3d

Again, T11 = a + (11 - 1)d  = a + 10d
=​ -3d + 10 d = 7d             [ Using (1)]

Also, T25 = a + (25 - 1)d =​ a + 24d = -3d + 24d = 21d             [ Using (1)]​
i.e., T25 = 3 ⨯ 7d = (3 ⨯ T11)
​Hence, 25th term is triple its 11th term.

#### Question 28:

The 8th term of an AP ia zero. Prove that its 38th term is triple its 18th term.                          [CBSE 2010]

Let a be the first term and d be the common difference of the AP. Then,

Now,

Hence, the 38th term of the AP is triple its 18th term.

#### Question 29:

The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference.       [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Now,

${a}_{5}+{a}_{7}=34$                (Given)

From (1) and (2), we get

$11-3d+5d=17\phantom{\rule{0ex}{0ex}}⇒2d=17-11=6\phantom{\rule{0ex}{0ex}}⇒d=3$

Hence, the common difference of the AP is 3.

#### Question 30:

The 9th term of an AP is −32 and the sum of its 11th and 13th terms is −94. Find the common difference of the AP.       [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Now,

${a}_{11}+{a}_{13}=-94$                (Given)

From (1) and (2), we get

$-32-8d+11d=-47\phantom{\rule{0ex}{0ex}}⇒3d=-47+32=-15\phantom{\rule{0ex}{0ex}}⇒d=-5$

Hence, the common difference of the AP is −5.

#### Question 31:

Determine the nth term of the AP whose 7th term is −1 and 16th term is 17.                      [CBSE 2014]

Let a be the first term and d be the common difference of the AP. Then,

Also,

From (1) and (2), we get

$-1-6d+15d=17\phantom{\rule{0ex}{0ex}}⇒9d=17+1=18\phantom{\rule{0ex}{0ex}}⇒d=2$

Putting d = 2 in (1), we get

$a+6×2=-1\phantom{\rule{0ex}{0ex}}⇒a=-1-12=-13$

Hence, the nth term of the AP is (2n − 15).

#### Question 32:

If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.                   [CBSE 2012]

Let a be the first term and d be the common difference of the AP. Then,

$4×{a}_{4}=18×{a}_{18}$      (Given)

$⇒a=-21d\phantom{\rule{0ex}{0ex}}⇒a+21d=0\phantom{\rule{0ex}{0ex}}⇒a+\left(22-1\right)d=0\phantom{\rule{0ex}{0ex}}⇒{a}_{22}=0$

Hence, the 22nd term of the AP is 0.

#### Question 33:

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

Let a be the first term and d be the common difference of the AP. Then,

$10×{a}_{10}=15×{a}_{15}$      (Given)

$⇒a+24d=0\phantom{\rule{0ex}{0ex}}⇒a+\left(25-1\right)d=0\phantom{\rule{0ex}{0ex}}⇒{a}_{25}=0$

Hence, the 25th term of the AP is 0.

#### Question 34:

Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.                                                                                                                                                                                                [CBSE 2012]

Let the common difference of the AP be d.

First term, a = 5

Now,
${a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}=\frac{1}{2}\left({a}_{5}+{a}_{6}+{a}_{7}+{a}_{8}\right)$               (Given)
$⇒a+\left(a+d\right)+\left(a+2d\right)+\left(a+3d\right)=\frac{1}{2}\left[\left(a+4d\right)+\left(a+5d\right)+\left(a+6d\right)+\left(a+7d\right)\right]$                 $\left[{a}_{n}=a+\left(n-1\right)d\right]$
$⇒4a+6d=\frac{1}{2}\left(4a+22d\right)\phantom{\rule{0ex}{0ex}}⇒8a+12d=4a+22d\phantom{\rule{0ex}{0ex}}⇒22d-12d=8a-4a\phantom{\rule{0ex}{0ex}}⇒10d=4a$

Hence, the common difference of the AP is 2.

#### Question 35:

The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.               [CBSE 2014]

Let a be the first term and d be the common difference of the AP. Then,

${a}_{2}+{a}_{7}=30$                     (Given)

Also,
${a}_{15}=2{a}_{8}-1$                   (Given)
$⇒a+14d=2\left(a+7d\right)-1\phantom{\rule{0ex}{0ex}}⇒a+14d=2a+14d-1\phantom{\rule{0ex}{0ex}}⇒-a=-1\phantom{\rule{0ex}{0ex}}⇒a=1$

Putting a = 1 in (1), we get
$2×1+7d=30\phantom{\rule{0ex}{0ex}}⇒7d=30-2=28\phantom{\rule{0ex}{0ex}}⇒d=4$

So,
${a}_{2}=a+d=1+4=5$
${a}_{3}=a+2d=1+2×4=9$,...

Hence, the AP is 1, 5, 9, 13,...

#### Question 36:

For what value of n, the nth terms of the arithmetic progressions 63, 65, 67,... and 3, 10, 17,... are equal?

Let the nth term of the given progressions be tn and Tn, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its nth term is given by
tn = a + (n - 1)
63 + (n - 1) ⨯ 2
61 + 2n

The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its nth term is given by
Tn = A + (n - 1)
3 + (n - 1) ⨯ 7
⇒ 7n - 4
Nowtn = ​Tn

⇒ 61 + 2n= 7n - 4​
65 = 5n
n = 13
Hence, the 13th terms of the AP's are the same.

#### Question 37:

The 17th term of an AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term.                   [CBSE 2012]

Let a be the first term and d be the common difference of the AP. Then,

${a}_{17}=2{a}_{8}+5$                (Given)

Also,
${a}_{11}=43$           (Given)

From (1) and (2), we get
$-5+2d+10d=43\phantom{\rule{0ex}{0ex}}⇒12d=43+5=48\phantom{\rule{0ex}{0ex}}⇒d=4$

Puting d = 4 in (1), we get
$a-2×4=-5\phantom{\rule{0ex}{0ex}}⇒a=-5+8=3$

Hence, the nth term of the AP is (4n − 1).

#### Question 38:

The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.                                 [CBSE 2013]

Let a be the first term and d be the common difference of the AP. Then,

${a}_{24}=2{a}_{10}$           (Given)

Now,

Hence, the 72nd term of the AP is 4 times its 15th term.

#### Question 39:

The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.                                                          [CBSE 2013]

Let a be the first term and d be the common difference of the AP. Then,

Also,

From (1) and (2), we get
$\frac{3d}{2}+8d=19\phantom{\rule{0ex}{0ex}}⇒\frac{3d+16d}{2}=19\phantom{\rule{0ex}{0ex}}⇒19d=38\phantom{\rule{0ex}{0ex}}⇒d=2$

Putting d = 2 in (1), we get
$2a=3×2=6\phantom{\rule{0ex}{0ex}}⇒a=3$

So,
${a}_{2}=a+d=3+2=5\phantom{\rule{0ex}{0ex}}{a}_{3}=a+2d=3+2×2=7,...$

Hence, the AP is 3, 5, 7, 9, ... .

#### Question 40:

If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.

In the given AP, let the first term be a and the common difference be d.
Then Tn = a + (n - 1)d ​
⇒ Tp = a + (p - 1)d = q               ...(i)
​

⇒ Tq = a + (q - 1)d = p              ...(ii)

On subtracting (i) from (ii), we get:
(q - p)d = (p - q)
⇒ d = -1
Putting d = -1 in (i), we get:
a = (pq - 1)

Thus, a = (p + q - 1) and d = -1
Now, Tp+q = a + (pq - 1)d
=​ (p + q - 1) + (p + q - 1)(-1)
= (p + q - 1) - (p + q - 1) = 0 ​

​Hence, the (p+q)th term is 0 (zero).

#### Question 41:

The first and last terms of an AP are a and l, respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a + 1).

In the given AP, first term = a and last term = l.
Let the common difference be d.
Then, nth term from the beginning is given by
Tn = a + (n - 1)d             ...(1)

Similarly, nth term from the end is given by
Tn = {l - (n - 1)d}            ...(2)
Adding (1) and (2), we get:
a + (n - 1)d + {l - (n - 1)d}​
= a + (n - 1)d + l - (n - 1)d
= a
+ l

​Hence, the sum of the nth term from the beginning and the nth term from the end is (a + l) .

#### Question 42:

How many two-digit numbers are divisible by 6?                           [CBSE 2011]

The two-digit numbers divisible by 6 are 12, 18, 24, ..., 96.

Clearly, these number are in AP.

Here, a = 12 and d = 18 − 12 = 6

Let this AP contains n terms. Then,

Hence, there are 15 two-digit numbers divisible by 6.

#### Question 43:

How many two-digit numbers are divisible by 3?                           [CBSE 2012]

The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.

Clearly, these number are in AP.

Here, a = 12 and d = 15 − 12 = 3

Let this AP contains n terms. Then,

$⇒n=30$

Hence, there are 30 two-digit numbers divisible by 3.

#### Question 44:

How many three-digit numbers are divisible by 9?                           [CBSE 2013]

The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.

Clearly, these number are in AP.

Here, a = 108 and d = 117 − 108 = 9

Let this AP contains n terms. Then,

Hence, there are 100 three-digit numbers divisible by 9.

#### Question 45:

How many numbers are there between 101 and 999, which are divisible by both 2 and 5?                           [CBSE 2014]

The numbers which are divisible by both 2 and 5 are divisible by 10 also.

Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, ..., 990.

Clearly, these number are in AP.

Here, a = 110 and d = 120 − 110 = 10

Let this AP contains n terms. Then,

Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

#### Question 46:

In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?

The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11.

Difference of rose plants between two consecutive rows = (41 - 43) = (39 - 41) = -2  [Constant]
So, the given progression is an AP.
Here, first term = 43
Common difference = -2
Last term = 11
Let n be the last term, then we have:
Tn = a + (n - 1)d
⇒ 11 = 43 + (n - 1)(-2)
⇒ 11 = 45 - 2n
⇒ 34 = 2n
⇒ n = 17
Hence, the 17th term is 11 or there are 17 rows in the flower bed.

#### Question 47:

A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes.

Let the amount of the first prize be ₹a.

Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.

Amount of the second prize = ₹(a − 200)

Amount of the third prize = ₹(a − 2 × 200) = ₹(a − 400)

Amount of the fourth prize = ₹(a − 3 × 200) = ₹(a − 600)

Now,

Total sum of the four prizes = ₹2,800

∴ ₹a + ₹(a − 200) + ₹(a − 400) + ₹(a − 600) = ₹2,800

⇒ 4− 1200 = 2800

⇒ 4a = 2800 + 1200 = 4000

a = 1000

∴ Amount of the first prize = ₹1,000

Amount of the second prize = ₹(1000 − 200) = ₹800

Amount of the third prize = ₹(1000 − 400) = ₹600

Amount of the fourth prize = ₹( 1000 − 600) = ₹400

Hence, the value of each of the prizes is ₹1,000, ₹800, ₹600 and ₹400.

#### Question 1:

Determine k so that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.                     [CBSE 2009C]

It is given that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.

∴ (4k − 6) − (3k − 2) = (k + 2) − (4k − 6)

⇒ 4− 6 − 3k + 2 = k + 2 − 4k + 6

− 4 = −3k + 8

k + 3k = 8 + 4

⇒ 4k = 12

k = 3

Hence, the value of k is 3.

#### Question 2:

Find the value of x for which the numbers (5x + 2), (4x − 1) and (x + 2) are in AP.

It is given that (5x + 2), (4x − 1) and (x + 2) are in AP.
$\therefore \left(4x-1\right)-\left(5x+2\right)=\left(x+2\right)-\left(4x-1\right)\phantom{\rule{0ex}{0ex}}⇒4x-1-5x-2=x+2-4x+1\phantom{\rule{0ex}{0ex}}⇒-x-3=-3x+3\phantom{\rule{0ex}{0ex}}⇒3x-x=3+3$
$⇒2x=6\phantom{\rule{0ex}{0ex}}⇒x=3$

Hence, the value of x is 3.

#### Question 3:

If (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.                        [CBSE 2014]

It is given that (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.

$⇒y=5$

Hence, the value of y is 5.

#### Question 4:

Write the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of an AP?

Since (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x+2) = (2x+3-2x
⇒ x - 2 = 3
x = 5​
∴ x = 5

#### Question 5:

Show that and ${\left(a+b\right)}^{2}$ are in AP.

The given numbers are and ${\left(a+b\right)}^{2}$.
Now,

$\left({a}^{2}+{b}^{2}\right)-{\left(a-b\right)}^{2}={a}^{2}+{b}^{2}-\left({a}^{2}-2ab+{b}^{2}\right)={a}^{2}+{b}^{2}-{a}^{2}+2ab-{b}^{2}=2ab$

${\left(a+b\right)}^{2}-\left({a}^{2}+{b}^{2}\right)={a}^{2}+2ab+{b}^{2}-{a}^{2}-{b}^{2}=2ab$

So,

Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.

#### Question 6:

Find three numbers in AP whose sum is 15 and product is 80.

Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 15
⇒ 3a = 15
⇒ a = 5​
Also, (a - d).a.(a + d)​ = 80
⇒ a(a2 - d2) = 80​
5 ​(25 - d2) = 80​   ​
d2 = 25​ - 16 =  9
d = $±$3
Thus, a = 5 and $±$3
Hence, the required numbers are (2, 5 and 8) or ( 8, 5 and 2).

#### Question 7:

The sum of three numbers in AP is 3 their product is −35. Find the numbers.

Let the required numbers be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1​
Also, (a - d).a.(a + d)​ = -35
⇒ a(a2 - d2) = -35​
1​.​(1​ - d2) = -35​ ​   ​
d2 = 36
d = $±$6

Thus, a = 1 and $±$6
Hence, the required numbers are ( -5, 1 and  7) or ( 7, 1 and -5).

#### Question 8:

Divide 24 in three parts such that they are in AP and their product is 440.

Let the required parts of 24 be (a - d), a and (a + d) such that they are in AP.
Then (a - d) + a + (a + d) = 24
⇒ 3a = 24
⇒ a =
8​
Also, (a - d).a.(a + d)​ = 440
⇒ a(a2 - d2) = 440​
⇒ 8​(64​​ -d2) = 440​    ​
d2 = 64 - 55 = 9
$±$3
Thus, a = 8 and $±$3
Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5).

#### Question 9:

The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.

Let the required terms be (a - d), a and (a + d).
Then (a - d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7​
Also, (a - d)2 + a2 + (a + d)2​ = 165
⇒ 3a2 + 2d2 = 165
​(3 ⨯ 49​​ + 2 d2) = 165   ​
⇒ 2d2 = 165 - 147 = 18
d2  = 9​
⇒ d = $±$3
Thus, a = 7 and $±$3
Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).

#### Question 10:

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

Let the required angles be (a - 15)o, (a - 5)o, (a + 5)o and (a + 15)o, as the common difference is 10 (given).
Then (a - 15)o + (a - 5)o + (a + 5)o + (a + 15)o = 360o
⇒ 4a = 360
⇒ a =
90

Hence, the required angles of a quadrilateral are

#### Question 12:

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth term is to the product of the second and the third term is 7 : 15.                                                                                                                                                                 [CBSE 2014]

Let the four parts in AP be (a − 3d), (ad), (a + d) and (a + 3d). Then,

Also,

$⇒960-135{d}^{2}=448-7{d}^{2}\phantom{\rule{0ex}{0ex}}⇒135{d}^{2}-7{d}^{2}=960-448\phantom{\rule{0ex}{0ex}}⇒128{d}^{2}=512\phantom{\rule{0ex}{0ex}}⇒{d}^{2}=4$
$⇒d=±2$

When a = 8 and d = 2,
$a-3d=8-3×2=8-6=2\phantom{\rule{0ex}{0ex}}a-d=8-2=6\phantom{\rule{0ex}{0ex}}a+d=8+2=10\phantom{\rule{0ex}{0ex}}a+3d=8+3×2=8+6=14$

When a = 8 and d = −2,
$a-3d=8-3×\left(-2\right)=8+6=14\phantom{\rule{0ex}{0ex}}a-d=8-\left(-2\right)=8+2=10\phantom{\rule{0ex}{0ex}}a+d=8-2=6\phantom{\rule{0ex}{0ex}}a+3d=8+3×\left(-2\right)=8-6=2$

Hence, the four parts are 2, 6, 10 and 14.

#### Question 13:

The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12, find the AP.                                                                                                                                                                                           [CBSE 2013C]

Let the first three terms of the AP be (ad), a and (a + d). Then,
$\left(a-d\right)+a+\left(a+d\right)=48\phantom{\rule{0ex}{0ex}}⇒3a=48\phantom{\rule{0ex}{0ex}}⇒a=16$

Now,

$⇒20d=180\phantom{\rule{0ex}{0ex}}⇒d=9$

When a = 16 and d = 9,
$a-d=16-9=7\phantom{\rule{0ex}{0ex}}a+d=16+9=25$

Hence, the first three terms of the AP are 7, 16 and 25.

#### Question 1:

The first three terms of an AP are respectively (3y − 1), (3y + 5) and (5y + 1), find the value of y.                                        [CBSE 2014]

The terms (3y − 1), (3y + 5) and (5y + 1) are in AP.
$\therefore \left(3y+5\right)-\left(3y-1\right)=\left(5y+1\right)-\left(3y+5\right)\phantom{\rule{0ex}{0ex}}⇒3y+5-3y+1=5y+1-3y-5\phantom{\rule{0ex}{0ex}}⇒6=2y-4\phantom{\rule{0ex}{0ex}}⇒2y=10$
$⇒y=5$

Hence, the value of y is 5.

#### Question 2:

If k, (2k − 1) and (2k + 1) are the three successive terms of an AP, find the value of k.                                        [CBSE 2014]

It is given that k, (2k − 1) and (2k + 1) are the three successive terms of an AP.
$\therefore \left(2k-1\right)-k=\left(2k+1\right)-\left(2k-1\right)\phantom{\rule{0ex}{0ex}}⇒k-1=2\phantom{\rule{0ex}{0ex}}⇒k=3$

Hence, the value of k is 3.

#### Question 3:

If 18, a, (b − 3) are in AP, then find the value of (2b).                                            [CBSE 2014]

It given that 18, a, (b − 3) are in AP.
$\therefore a-18=\left(b-3\right)-a\phantom{\rule{0ex}{0ex}}⇒a+a-b=18-3\phantom{\rule{0ex}{0ex}}⇒2a-b=15$

Hence, the required value is 15.

#### Question 4:

If the numbers a, 9, b, 25 form an AP, find a and b.                              [CBSE 2014]

It is given that the numbers a, 9, b, 25 form an AP.
$\therefore 9-a=b-9=25-b$
So,
$b-9=25-b\phantom{\rule{0ex}{0ex}}⇒2b=34\phantom{\rule{0ex}{0ex}}⇒b=17$
Also,

Hence, the required values of a and b are 1 and 17, respectively.

#### Question 5:

If the numbers (2n − 1), (3n + 2) and (6− 1) are in AP, find the value of n and the numbers.                          [CBSE 2013C]

It is given that the numbers (2n − 1), (3n + 2) and (6− 1) are in AP.
$\therefore \left(3n+2\right)-\left(2n-1\right)=\left(6n-1\right)-\left(3n+2\right)\phantom{\rule{0ex}{0ex}}⇒3n+2-2n+1=6n-1-3n-2\phantom{\rule{0ex}{0ex}}⇒n+3=3n-3\phantom{\rule{0ex}{0ex}}⇒2n=6$
$⇒n=3$

When n = 3,
$2n-1=2×3-1=6-1=5\phantom{\rule{0ex}{0ex}}3n+2=3×3+2=9+2=11\phantom{\rule{0ex}{0ex}}6n-1=6×3-1=18-1=17$

Hence, the required value of n is 3 and the numbers are 5, 11 and 17.

#### Question 6:

How many three-digit natural numbers are divisible by 7?                                                         [CBSE 2013]

The three-digit natural numbers divisible by 7 are 105, 112, 119, ..., 994.

Clearly, these number are in AP.

Here, a = 105 and d = 112 − 105 = 7

Let this AP contains n terms. Then,

Hence, there are 128 three-digit numbers divisible by 7.

#### Question 7:

How many three-digit natural numbers are divisible by 9?                                                         [CBSE 2013]

The three-digit natural numbers divisible by 9 are 108, 117, 126, ..., 999.

Clearly, these number are in AP.

Here, a = 108 and d = 117 − 108 = 9

Let this AP contains n terms. Then,

Hence, there are 100 three-digit numbers divisible by 9.

#### Question 8:

If the sum of first m terms of an AP is (2m2 + 3m) then what is its second term?                                [CBSE 2011]

Let Sm denotes the sum of first m terms of the AP.

Now,
mth term of the AP, am = Sm − S− 1

Putting m = 2, we get
${a}_{2}=4×2+1=9$

Hence, the second term of the AP is 9.

#### Question 9:

What is the sum of first n terms of the AP a, 3a, 5a, ... .                                [CBSE 2012]

The given AP is a, 3a, 5a, ... .

Here,

First term, A = a

Common difference, D = 3a a = 2a

∴ Sum of first n terms, Sn

Hence, the required sum is an2.

#### Question 10:

What is the 5th term from the end of the AP 2, 7, 12, ..., 47?                       [CBSE 2011]

The given AP is 2, 7, 12, ..., 47.

Let us re-write the given AP in reverse order i.e. 47, 42, ..., 12, 7, 2.

Now, the 5th term from the end of the given AP is equal to the 5th term from beginning of the AP 47, 42, ..., 12, 7, 2.

Consider the AP 47, 42, ..., 12, 7, 2.

Here, a = 47 and d = 42 − 47 = −5

5th term of this AP

= 47 + (5 − 1) × (−5)

= 47 − 20

= 27

Hence, the 5th term from the end of the given AP is 27.

#### Question 11:

If an denotes the nth term of the AP 2, 7, 12, 17, ... , find the value of (a30 − a20).                           [CBSE 2011]

The given AP is 2, 7, 12, 17, ... .

Here, a = 2 and d = 7 − 2 = 5

Hence, the required value is 50.

#### Question 12:

The nth term of an AP is (3n + 5). Find its common difference.

We have:
Tn = (3n + 5)
Common difference = T2 - T1
​T1 = 3 ⨯ 1 + 5 = 8
T2 = ​3 ⨯ 2 + 5 = 11
d = 11 - 8 = 3
Hence, the common difference is 3.

#### Question 13:

The nth term of an AP is (7 − 4n). Find its common difference.

We have:
Tn = (7 - 4n)
Common difference = T2 - T1
​T= 7 - 4 ⨯ 1 = 3
T2 = ​7 - 4 ⨯ 2 = -1
d = -1 - 3 = -4
Hence, the common difference is -4.

#### Question 14:

Write the next term of the AP

#### Question 15:

Write the next term of the AP

:

#### Question 16:

Which term of the AP 21, 18, 15,... is zero?

In the given AP, first term, a = 21 and common difference, d = (18 - 21) = -3
Let's its nth term be 0.
Then
Tn = 0
⇒
a + (n - 1)d = 0
​ ⇒ 21 + (n - 1) ⨯ (-3) = 0
⇒ 24 - 3n = 0
⇒ 3n = 24
⇒ n = 8
Hence, the 8th term of the given AP is 0.

#### Question 17:

Find the sum of first n natural numbers.

The first n natural numbers are 1, 2, 3, 4, 5, ..., n.

Here, a = 1 and d = (2 - 1) = 1

#### Question 18:

Find the sum of first n even natural numbers.

The first n even natural numbers are 2 ,4, 6, 8, 10, ..., n.

Here, = 2 and d = (4 - 2) = 2

Hence, the required sum is n(n+1).

#### Question 19:

The first term of an AP is p and its common difference is q. Find its 10th term.

Here, a = p and d = q
Now, Tn = a + (n - 1)d
Tn = p + (n - 1)q
∴ ​T10p + 9q

#### Question 20:

If $\frac{4}{5},a,$ 2 are three consecutive terms of an AP, find the value of a.

If  $\frac{4}{5}$, a and 2 are three consecutive terms of an AP, then we have:
a$\frac{4}{5}$ = 2 - a
⇒ 2a  = 2 + $\frac{4}{5}$
2a$\frac{14}{5}$
a = $\frac{7}{5}$

#### Question 21:

If $\left(2p+1\right),13,\left(5p-3\right)$  are in AP, find the value of p.

Let (2p+1), 13,(5p-3) be three consecutive terms of an AP.
Then
13 - (2p+1) =(5p 3) - 13
⇒ 7p = 28
⇒ p = 4​
∴ When p = 4,
(2p+1), 13 and(5p-3) form three consecutive terms of an AP.

#### Question 22:

If $\left(2p-1\right),$ 7 and 3p are in AP, find the value of p.

Let (2p1), 7 and 3p be three consecutive terms of an AP.
Then
7  (21) = 3p  7
⇒ 5p = 15
⇒ p = 3​
∴ When p = 3,
(2p1), 7 and 3p form three consecutive terms of an AP.

#### Question 23:

If the sum of first p terms of an AP is (ap2 + bp), find its common difference.                        [CBSE 2010]

Let Sp denotes the sum of first p terms of the AP.

Now,
pth term of the AP, ap = Sp − Sp − 1
$=\left(a{p}^{2}+bp\right)-\left[a{p}^{2}-\left(2a-b\right)p+\left(a-b\right)\right]\phantom{\rule{0ex}{0ex}}=a{p}^{2}+bp-a{p}^{2}+\left(2a-b\right)p-\left(a-b\right)\phantom{\rule{0ex}{0ex}}=2ap-\left(a-b\right)$

Let d be the common difference of the AP.

Hence, the common difference of the AP is 2a.

#### Question 24:

If the sum of first n terms is (3n2 + 5n), find its common difference.

Let Sn denotes the sum of first n terms of the AP.

Now,
nth term of the AP, an = Sn − Sn − 1
$=\left(3{n}^{2}+5n\right)-\left(3{n}^{2}-n-2\right)\phantom{\rule{0ex}{0ex}}=6n+2$

Let d be the common difference of the AP.

Hence, the common difference of the AP is 6.

#### Question 25:

Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.                         [CBSE 2011]

Let a be the first term and d be the common difference of the AP. Then,

Now,
${a}_{6}+{a}_{13}=40$                (Given)

From (1) and (2), we get
$2\left(9-3d\right)+17d=40\phantom{\rule{0ex}{0ex}}⇒18-6d+17d=40\phantom{\rule{0ex}{0ex}}⇒11d=40-18=22\phantom{\rule{0ex}{0ex}}⇒d=2$

Putting d = 2 in (1), we get
$a+3×2=9\phantom{\rule{0ex}{0ex}}⇒a=9-6=3$

Hence, the AP is 3, 5, 7, 9, 11, ... .

#### Question 1:

Find the sum of each of the following APs:

(i) 2, 7, 12, 17, ... to 19 terms
(ii) 9, 7, 5, 3, ... to 14 terms
(iii) −37, −33, −29, ... to 12 terms
(iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ to 11 terms
(v) 0.6, 1.7, 2.8, ... to 100 terms

(i) The given AP is 2, 7, 12, 17, ... .

Here, a = 2 and d = 7 − 2 = 5

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have

(ii) The given AP is 9, 7, 5, 3, ... .

Here, a = 9 and d = 7 − 9 = −2

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have

(iii) The given AP is −37, −33, −29, ... .

Here, a = −37 and d = −33 − (−37) = −33 + 37 = 4

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have

(iv) The given AP is $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ .

Here, a$\frac{1}{15}$ and d = $\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have

(v) The given AP is 0.6, 1.7, 2.8, ... .

Here, a = 0.6 and d = 1.7 − 0.6 = 1.1

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have

#### Question 2:

Find the sum of each of the following arithmetic series:

(i) $7+10\frac{1}{2}+14+...+84$
(ii) 34 + 32 + 30 + ... + 10
(iii) (−5) + (−8) + (−11) + ... + (−230)

(i) The given arithmetic series is $7+10\frac{1}{2}+14+...+84$.

Here, a = 7, d = $10\frac{1}{2}-7=\frac{21}{2}-7=\frac{21-14}{2}=\frac{7}{2}$ and l = 84.

Let the given series contain n terms. Then,

$⇒n=\frac{161}{7}=23$
∴ Required sum = $\frac{23}{2}×\left(7+84\right)$                  $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{23}{2}×91\phantom{\rule{0ex}{0ex}}=\frac{2093}{2}\phantom{\rule{0ex}{0ex}}=1046\frac{1}{2}$

(ii) The given arithmetic series is 34 + 32 + 30 + ... + 10.

Here, a = 34, d = 32 − 34 = −2  and l = 10.

Let the given series contain n terms. Then,

$⇒n=13$
∴ Required sum = $\frac{13}{2}×\left(34+10\right)$                  $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{13}{2}×44\phantom{\rule{0ex}{0ex}}=286$

(iii) The given arithmetic series is (−5) + (−8) + (−11) + ... + (−230).

Here, a = −5, d = −8 − (−5) = −8 + 5 = −3  and l = −230.

Let the given series contain n terms. Then,

$⇒n=76$
∴ Required sum = $\frac{76}{2}×\left[\left(-5\right)+\left(-230\right)\right]$                  $\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]$
$=\frac{76}{2}×\left(-235\right)\phantom{\rule{0ex}{0ex}}=-8930$

#### Question 3:

Find the sum of first n terms of an AP whose nth term is (5 − 6n). Hence, find the sum of its first 20 terms.

Let an be the nth term of the AP.

∴ an = 5 − 6n

Putting n = 1, we get

First term, a = a1 = 5 − 6 × 1 = −1

Putting n = 2, we get

a2 = 5 − 6 × 2 = −7

Let d be the common difference of the AP.

d = ${a}_{2}-{a}_{1}=-7-\left(-1\right)=-7+1=-6$

Sum of first n terms of the AP, Sn

Putting n = 20, we get

${S}_{20}=2×20-3×{20}^{2}=40-1200=-1160$

#### Question 4:

The sum of the first n terms of an AP is (3n2 + 6n). Find the nth term and the 15th term of this AP.                           [CBSE 2014]

Let Sn denotes the sum of first n terms of the AP.

nth term of the AP, an
$={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}=\left(3{n}^{2}+6n\right)-\left(3{n}^{2}-3\right)\phantom{\rule{0ex}{0ex}}=6n+3$

Putting n = 15, we get

${a}_{15}=6×15+3=90+3=93$

Hence, the nth term is (6n + 3) and 15th term is 93.

#### Question 5:

If the sum of the first n terms of an AP is given by Sn = (3n2 − n), find its (i) nth term, (ii) first term, and (iii) common difference.

Given: Sn = (3n2n)               ...(i)
Replacing n by (n - 1) in (i), we get:
Sn-1 = 3(n - 1)2 - (n - 1)
= 3(n2 - 2n + 1) - + 1
= 3n2 - 7n + 4
(i) Now,​ Tn = ( Sn - Sn-1)
​    =
(3n2 - n) - (3n2 - 7n + 4 ) = 6n - 4
∴ nth term, Tn = (
6n - 4)                ...(ii)

(ii) Putting n = 1 in (ii), we get:
T1= (6 ⨯ 1) - 4 = 2

(iii) Putting n = 2 in (ii), we get:
T2= (6 ⨯ 2) - 4 = 8
∴ Common difference, d = T2 - T1 = 8 - 2 = 6

#### Question 6:

The sum of n terms of an AP is 49 and that of first 17 terms is 289, find the sum of first n terms.

Putting n = 20 in (ii), we get:
T20 = (5 ⨯ 20) - 1 = 99
Hence, the 20th term is 99.

#### Question 7:

The sum of the first n terms of an AP is $\left(\frac{3{n}^{2}}{2}+\frac{5n}{2}\right)$. Find its nth term and the 25th term.                      [CBSE 2006C]

Let Sn denotes the sum of first n terms of the AP.

nth term of the AP, an
$={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}=\left(\frac{3{n}^{2}+5n}{2}\right)-\left(\frac{3{n}^{2}-n-2}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{6n+2}{2}\phantom{\rule{0ex}{0ex}}=3n+1$

Putting n = 25, we get

${a}_{25}=3×25+1=75+1=76$

Hence, the nth term is (3n + 1) and 25th term is 76.

#### Question 8:

How many terms of the AP 21, 18, 15, ... must be added to get the sum 0?

The given AP is 21, 18, 15, ... .

Here, a = 21 and d = 18 − 21 = −3

Let the required number of terms be n. Then,

n = 15                 (Number of terms cannot be zero)

Hence, the required number of terms is 15.

#### Question 9:

How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

The given AP is 9, 17, 25, ... .

Here, a = 9 and d = 17 − 9 = 8

Let the required number of terms be n. Then,

$⇒n\left(5+4n\right)=636\phantom{\rule{0ex}{0ex}}⇒4{n}^{2}+5n-636=0\phantom{\rule{0ex}{0ex}}⇒4{n}^{2}-48n+53n-636=0\phantom{\rule{0ex}{0ex}}⇒4n\left(n-12\right)+53\left(n-12\right)=0$

n = 12                    (Number of terms cannot negative)

Hence, the required number of terms is 12.

#### Question 10:

How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.

The given AP is 63, 60, 57, 54, ... .

Here, a = 63 and d = 60 − 63 = −3

Let the required number of terms be n. Then,

$⇒3{n}^{2}-129n+1386=0\phantom{\rule{0ex}{0ex}}⇒3{n}^{2}-66n-63n+1386=0\phantom{\rule{0ex}{0ex}}⇒3n\left(n-22\right)-63\left(n-22\right)=0$

So, the sum of 21 terms as well as that of 22 terms is 693. This is because the 22nd term of the AP is 0.

${a}_{22}=63+\left(22-1\right)×\left(-3\right)=63-63=0$

Hence, the required number of terms is 21 or 22.

#### Question 11:

How many terms of the AP 20, $19\frac{1}{3}$, $18\frac{2}{3}$, ... must be taken so that their sum is 300? Explain the double answer.

The given AP is 20, $19\frac{1}{3}$, $18\frac{2}{3}$, ... .

Here, a = 20 and d = $19\frac{1}{3}-20=\frac{58}{3}-20=\frac{58-60}{3}=-\frac{2}{3}$

Let the required number of terms be n. Then,

$⇒122n-2{n}^{2}=1800\phantom{\rule{0ex}{0ex}}⇒2{n}^{2}-122n+1800=0\phantom{\rule{0ex}{0ex}}⇒2{n}^{2}-50n-72n+1800=0\phantom{\rule{0ex}{0ex}}⇒2n\left(n-25\right)-72\left(n-25\right)=0$

So, the sum of first 25 terms as well as that of first 36 terms is 300. This is because the sum of all terms from 26th to 36th is 0.

Hence, the required number of terms is 25 or 36.

#### Question 12:

Find the sum of all odd numbers between 0 and 50.

All odd numbers between 0 and 50 are 1, 3, 5, 7, ..., 49.
This is an AP in which a = 1, d = (3 - 1) = 2 and l = 49.
Let the number of terms be n.
Then, Tn = 49
a + (n - 1)d = 49
⇒ 1 + (n - 1​) ⨯ 2 = 49
⇒ 2n50
⇒ n = 25

∴ Required sum = $\frac{n}{2}\left(a+l\right)$

=
Hence, the required sum is 625.

#### Question 13:

Find the sum of all natural numbers between 200 and 400 which are divisible by 7.                 [CBSE 2012]

Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, ..., 399.

This is an AP with a = 203, d = 7 and l = 399.

Suppose there are n terms in the AP. Then,

$⇒n=29$

∴ Required sum =
$=\frac{29}{2}×602\phantom{\rule{0ex}{0ex}}=8729$

Hence, the required sum is 8729.

#### Question 14:

Find the sum of first forty positive integers divisible by 6.                      [CBSE 2012]

The positive integers divisible by 6 are 6, 12, 18, ... .

This is an AP with a = 6 and d = 6.

Also, n = 40           (Given)

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the required sum is 4920.

#### Question 15:

Find the sum of first 15 multiples of 8.

The first 15 multiples of 8 are 8, 16, 24, 32,...
This is an AP in which a = 8, d = (16 - 8) = 8 and n = 15.
Thus, we have:

∴ Required sum = $\frac{n}{2}\left(a+l\right)$

=
Hence, the required sum is 960.

#### Question 16:

Find the sum of all multiples of 9 lying between 300 and 700.

The multiples of 9 lying between 300 and 700 are 306, 315, ..., 693.

This is an AP with a = 306, d = 9 and l = 693.

Suppose there are n terms in the AP. Then,

$⇒n=44$

∴ Required sum
$=22×999\phantom{\rule{0ex}{0ex}}=21978$

Hence, the required sum is 21978.

#### Question 17:

Find the sum of all three-digit natural numbers which are divisible by 13.

All three-digit numbers which are divisible by 13 are 104, 117, 130, 143,..., 988.

This is an AP in which a = 104, d = (117 - 104) = 13 and l = 988
Let the number of terms be n.
Then Tn = 988
a + (n - 1)d = 988
⇒ 104 + (n -1​) ⨯​ 13 = 988
⇒ 13n = 897
⇒ n = 69

∴ Required sum = $\frac{n}{2}\left(a+l\right)$
=
Hence, the required sum is 37674.

#### Question 18:

Find the sum of first 100 even natural numbers which are divisible by 5.

The first few even natural numbers which are divisible by 5 are 10, 20, 30, 40, ...
This is an AP in which a = 10, d = (20 − 10) = 10 and n = 100
The sum of n terms of an AP is given by

Hence, the sum of the first hundred even natural numbers which are divisible by 5 is 50500.

#### Question 19:

Find the sum of the following:
.

On simplifying the given series, we get:

#### Question 20:

In an AP, it is given that S5 + S7 = 167 and S10 = 235, then find the AP, where Sn denotes the sum of its first n terms.                   [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Also,

${S}_{10}=235\phantom{\rule{0ex}{0ex}}⇒\frac{10}{2}\left(2a+9d\right)=235\phantom{\rule{0ex}{0ex}}⇒5\left(2a+9d\right)=235\phantom{\rule{0ex}{0ex}}⇒2a+9d=47$

Multiplying both sides by 6, we get

Subtracting (1) from (2), we get

$12a+54d-12a-31d=282-167\phantom{\rule{0ex}{0ex}}⇒23d=115\phantom{\rule{0ex}{0ex}}⇒d=5$

Putting d = 5 in (1), we get

$12a+31×5=167\phantom{\rule{0ex}{0ex}}⇒12a+155=167\phantom{\rule{0ex}{0ex}}⇒12a=167-155=12\phantom{\rule{0ex}{0ex}}⇒a=1$

Hence, the AP is 1, 6, 11, 16, ... .

#### Question 21:

In an AP the first term is 2, the last term is 29 and sum of all the terms is 155. Find the common difference of the AP.

Here, a = 2, l = 29 and Sn = 155
Let d be the common difference of the given AP and n be the total number of terms.
Then Tn = 29
⇒ a + (- 1)d = 29
⇒ 2 + (- 1)d = 29​                ...(i)

The sum of terms of an AP is given by

Putting the value of n in (i), we get:
⇒ 2 + 9d = 29
⇒ 9d = 27
⇒ d = 3
Thus, the common difference of the given AP is 3.

#### Question 22:

In an AP, the first term is −4, the last term is 29 and the sum of all its terms is 150. Find its common difference.                    [CBSE 2011]

Suppose there are n terms in the AP.

Here, a = −4, l = 29 and Sn = 150

Thus, the AP contains 12 terms.

Let d be the common difference of the AP.

Hence, the common difference of the AP is 3.

#### Question 23:

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Suppose there are n terms in the AP.

Here, a = 17, d = 9 and l = 350

$⇒n=38$

Thus, there are 38 terms in the AP.

Hence, the required sum is 6973.

#### Question 24:

The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms.                                                                                                                                                                                      [CBSE 2012, 14]

Suppose there are n terms in the AP.

Here, a = 5, l = 45 and Sn = 400

Thus, there are 16 terms in the AP.

Let d be the common difference of the AP.

Hence, the common difference of the AP is $\frac{8}{3}$.

#### Question 25:

In an AP the first term is 22, nth term is −11 and sum to first nth terms is 66. Find n and d, the common difference.

Here, a = 22, Tn = -11 and Sn = 66
Let d be the common difference of the given AP.
Then Tn = -11
⇒ a + (n - 1)d = 22 + (n - 1)d = -11
⇒ (n - 1)d = -33        ...(i)

The sum of n terms of an AP is given by
[Substituting the value of (n - 1)d from (i)]
Putting the value of n in (i), we get:
11d = -33
d = -3
Thus, n = 12 and d = -3

#### Question 26:

The 12th term of an AP is −13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.                                [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Also,

Solving (1) and (2), we get

$2\left(-13-11d\right)+3d=12\phantom{\rule{0ex}{0ex}}⇒-26-22d+3d=12\phantom{\rule{0ex}{0ex}}⇒-19d=12+26=38\phantom{\rule{0ex}{0ex}}⇒d=-2$

Putting d = −2 in (1), we get

$a+11×\left(-2\right)=-13\phantom{\rule{0ex}{0ex}}⇒a=-13+22=9$

∴ Sum of its first 10 terms, S10

$=\frac{10}{2}\left[2×9+\left(10-1\right)×\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=5×\left(18-18\right)\phantom{\rule{0ex}{0ex}}=5×0\phantom{\rule{0ex}{0ex}}=0$

Hence, the required sum is 0.

#### Question 27:

The sum of the first 7 terms of an AP is 182. If its 4th and 17th terms are in the ratio 1 : 5, find the AP.                     [CBSE 2014]

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

Also,

Solving (1) and (2), we get

$a+3×4a=26\phantom{\rule{0ex}{0ex}}⇒13a=26\phantom{\rule{0ex}{0ex}}⇒a=2$

Putting a = 2 in (2), we get

$d=4×2=8$

Hence, the required AP is 2, 10, 18, 26, ... .

#### Question 28:

The first and last terms of an AP are 4 and 81 respectively. If the common difference is 7, how many terms are there in the AP and what is their sum?

Here, a = 4, d = 7 and l = 81
Let the nth term be 81.
Then Tn = 81
⇒ a + (n - 1)d = 4 + (n - 1)7 =  81
⇒ (n - 1)7 = 77
⇒ (n - 1) = 11
n = 12
Thus, there are 12 terms in the AP.

The sum of n terms of an AP is given by

Thus, the required sum is 510.

#### Question 29:

The sum of first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.

Let a be the first term and d be the common difference of the given AP.
Then we have:

However, S7 = 49 and S17 = 289
Now, 7[a + 3d] = 49
⇒ a + 3d = 7           ...(i)
Also, 17[a + 8d] = 289
​⇒ a + 8d = 17           ...(ii)

Subtracting (i) from (ii), we get:

5d = 10
⇒ d = 2

Putting d = 2 in (i), we get:
a + 6 = 7
⇒ a = 1
Thus, a = 1 and d = 2

∴ Sum of n terms of AP =

#### Question 30:

Two APs have the same common difference. If the first terms of these APs be 3 and 8 respectively, find the difference between the sums of their first 50 terms.                                                                                                                                                                       [CBSE 2011]

Let a1 and a2 be the first terms of the two APs.

Here, a1 = 8 and a2 = 3

Suppose d be the common difference of the two APs.

Let ${S}_{50}$ and ${S}_{50}^{\text{'}}$ denote the sums of their first 50 terms.

Hence, the required difference between the two sums is 250.

#### Question 31:

The sum of first 10 terms of an AP is −150 and the sum of its next 10 terms is −550. Find the AP.                                     [CBSE 2010]

Let a be the first term and d be the common difference of the AP. Then,

It is given that the sum of its next 10 terms is −550.

Now,

S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = −150 + (−550) = −700

Subtracting (1) from (2), we get

$\left(2a+19d\right)-\left(2a+9d\right)=-70-\left(-30\right)\phantom{\rule{0ex}{0ex}}⇒10d=-40\phantom{\rule{0ex}{0ex}}⇒d=-4$

Putting d = −4 in (1), we get

$2a+9×\left(-4\right)=-30\phantom{\rule{0ex}{0ex}}⇒2a=-30+36=6\phantom{\rule{0ex}{0ex}}⇒a=3$

Hence, the required AP is 3, −1, −5, −9, ... .

#### Question 32:

The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16, find the sum of its first 10 terms.                                [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Also,

Solving (1) and (2), we get

$a+3a=16\phantom{\rule{0ex}{0ex}}⇒4a=16\phantom{\rule{0ex}{0ex}}⇒a=4$

Putting a = 4 in (1), we get

$4d=3×4=12\phantom{\rule{0ex}{0ex}}⇒d=3$

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the required sum is 175.

#### Question 33:

The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.                                [CBSE 2015]

Let a be the first term and d be the common difference of the AP. Then,

Also,

Solving (1) and (2), we get

$a+9×\frac{4a}{5}=41\phantom{\rule{0ex}{0ex}}⇒\frac{5a+36a}{5}=41\phantom{\rule{0ex}{0ex}}⇒\frac{41a}{5}=41\phantom{\rule{0ex}{0ex}}⇒a=5$

Putting a = 5 in (1), we get

$5d=4×5=20\phantom{\rule{0ex}{0ex}}⇒d=4$

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the required sum is 495.

#### Question 34:

An AP 5, 12, 19, ... has 50 terms. Find its last term. Hence, find the sum of its last 15 terms.                                             [CBSE 2015]

The given AP is 5, 12, 19, ... .

Here, a = 5, d = 12 − 5 = 7 and n = 50.

Since there are 50 terms in the AP, so the last term of the AP is a50.

Thus, the last term of the AP is 348.

Now,

Sum of the last 15 terms of the AP

$=\frac{17650-8680}{2}\phantom{\rule{0ex}{0ex}}=\frac{8970}{2}\phantom{\rule{0ex}{0ex}}=4485$

Hence, the required sum is 4485.

#### Question 35:

An AP 8, 10, 12, ... has 60 terms. Find its last term. Hence, find the sum of its last 10 terms.                                             [CBSE 2015]

The given AP is 8, 10, 12, ... .

Here, a = 8, d = 10 − 8 = 2 and n = 60

Since there are 60 terms in the AP, so the last term of the AP is a60.

Thus, the last term of the AP is 126.

Now,

Sum of the last 10 terms of the AP

$=4020-2850\phantom{\rule{0ex}{0ex}}=1170$

Hence, the required sum is 1170.

#### Question 36:

The sum of the 4th and the 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms.                                                                                                                                                                                             [CBSE 2013C]

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

Also,

Subtracting (1) from (2), we get

$\left(a+7d\right)-\left(a+5d\right)=22-12\phantom{\rule{0ex}{0ex}}⇒2d=10\phantom{\rule{0ex}{0ex}}⇒d=5$

Putting d = 5 in (1), we get

$a+5×5=12\phantom{\rule{0ex}{0ex}}⇒a=12-25=-13$

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the required sum is 95.

#### Question 37:

The sum of first m terms of an AP is (4m2m). If its nth term is 107, find the value of n. Also, find the 21st term of this AP.                                                                                                                                                                                                       [CBSE 2013]

Let Sm denote the sum of the first m terms of the AP. Then,

Suppose am denote the mth term of the AP.

Now,

Thus, the value of n is 14.

Putting m = 21 in (1), we get

${a}_{21}=8×21-5=168-5=163$

Hence, the 21st term of the AP is 163.

#### Question 38:

The sum of first q terms of an AP is (63q − 3q2). If its pth term is −60, find the value of p. Also, find the 11th term of its AP.                                                                                                                                                                                                      [CBSE 2013]

Let Sq denote the sum of the first q terms of the AP. Then,

Suppose aq denote the qth term of the AP.

Now,

Thus, the value of p is 21.

Putting q = 11 in (1), we get

${a}_{11}=-6×11+66=-66+66=0$

Hence, the 11th term of the AP is 0.

#### Question 39:

Find the number of terms of the AP −12, −9, −6, ..., 21. If 1 is added to each term of this AP then find the sum of all terms of the AP thus obtained.                                                                                                                                                                                       [CBSE 2013]

The given AP is −12, −9, −6, ..., 21.

Here, a = −12, d = −9 − (−12) = −9 + 12 = 3 and l = 21

Suppose there are n terms in the AP.

$⇒n=12$

Thus, there are 12 terms in the AP.

If 1 is added to each term of the AP, then the new AP so obtained is −11, −8, −5, ..., 22.

Here, first term, A = −11; last term, L = 22 and n = 12

∴ Sum of the terms of this AP

$=6×11\phantom{\rule{0ex}{0ex}}=66$

Hence, the required sum is 66.

#### Question 40:

Sum of the first 14 terms of an AP is 1505 and its first term is 10. Find its 25th term.                                [CBSE 2012]

Let d be the common difference of the AP.

Here, a = 10 and n = 14
Now,

$⇒13d=215-20=195\phantom{\rule{0ex}{0ex}}⇒d=15$

∴ 25th term of the AP, a25

Hence, the required term is 370.

#### Question 41:

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Let a be the first term and d be the common difference of the AP. Then,

$d={a}_{3}-{a}_{2}=18-14=4$

Now,

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the required sum is 5610.

#### Question 42:

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by students. Which value is shown in the question?                                      [CBSE 2014]

Number of trees planted by the students of each section of class 1 = 2

There are two sections of class 1.

∴ Number of trees planted by the students of class 1 = 2 × 2 = 4

Number of trees planted by the students of each section of class 2 = 4

There are two sections of class 2.

∴ Number of trees planted by the students of class 2 = 2 × 4 = 8

Similarly,

Number of trees planted by the students of class 3 = 2 × 6 = 12

So, the number of trees planted by the students of differents classes are 4, 8, 12, ... .

∴ Total number of trees planted by the students = 4 + 8 + 12 + ... up to 12 terms

This series is an arithmetic series.

Here, a = 4, d = 8 − 4 = 4 and n = 12

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the total number of trees planted by the students is 312.

The values shown in the question are social responsibility and awareness for conserving nature.

#### Question 43:

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are 10 potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and he continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Distance covered by the competitor to pick and drop the first potato = 2 × 5 m = 10 m

Distance covered by the competitor to pick and drop the second potato = 2 × (5 + 3) m = 2 × 8 m = 16 m

Distance covered by the competitor to pick and drop the third potato = 2 × (5 + 3 + 3) m = 2 × 11 m = 22 m and so on.

∴ Total distance covered by the competitor = 10 m + 16 m + 22 m + ... up to 10 terms

This is an arithmetic series.

Here, a = 10, d = 16 − 10 = 6 and n = 10

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the total distance the competitor has to run is 370 m.

#### Question 44:

There are 25 trees at equal distances of 5 m in a line with a water tank, the distance of the water tank from the nearest tree being 10 m. A gardener waters all the trees separately, starting from the water tank and returning back to the water tank after watering each tree to get water for the next. Find the total distance covered by the gardener in order to water all the trees.

Distance covered by the gardener to water the first tree and return to the water tank = 10 m + 10 m = 20 m

Distance covered by the gardener to water the second tree and return to the water tank = 15 m + 15 m = 30 m

Distance covered by the gardener to water the third tree and return to the water tank = 20 m + 20 m = 40 m and so on.

∴ Total distance covered by the gardener to water all the trees = 20 m + 30 m + 40 m + ... up to 25 terms

This series is an arithmetic series.

Here, a = 20, d = 30 − 20 = 10 and n = 25

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the total distance covered by the gardener to water all the trees 3500 m.

#### Question 45:

A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each prize.

Let the value of the first prize be ₹a.

Since the value of each prize is ₹20 less than its preceding prize, so the values of the prizes are in AP with common difference −₹20.

d = −₹20

Number of cash prizes to be given to the students, n = 7

Total sum of the prizes, S7 = ₹700

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

${S}_{7}=\frac{7}{2}\left[2a+\left(7-1\right)×\left(-20\right)\right]=700\phantom{\rule{0ex}{0ex}}⇒\frac{7}{2}\left(2a-120\right)=700\phantom{\rule{0ex}{0ex}}⇒7a-420=700\phantom{\rule{0ex}{0ex}}⇒7a=700+420=1120$
$⇒a=160$

Thus, the value of the first prize is ₹160.

Hence, the value of each prize is ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.

#### Question 46:

A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more than he did in the preceding month. How much did he save in the first month?

Let the money saved by the man in the first month be ₹a.

It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.

d = ₹100

Number of months, n = 10

Sum of money saved in 10 months, S10 = ₹33,000

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

${S}_{10}=\frac{10}{2}\left[2a+\left(10-1\right)×100\right]=33000\phantom{\rule{0ex}{0ex}}⇒5\left(2a+900\right)=33000\phantom{\rule{0ex}{0ex}}⇒2a+900=6600\phantom{\rule{0ex}{0ex}}⇒2a=6600-900=5700$
$⇒a=2850$

Hence, the money saved by the man in the first month is ₹2,850.

#### Question 47:

A man arranges to pay off a debt of ₹36000 by 40 monthly 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.

Let the value of the first instalment be ₹a.

Since the monthly instalments form an arithmetic series, so let us suppose the man increases the value of each instalment by ₹d every month.

∴ Common difference of the arithmetic series = ₹d

Amount paid in 30 instalments = ₹36,000 − $\frac{1}{3}$ × ₹36,000 = ₹36,000 − ₹12,000 = ₹24,000

Let Sn denote the total amount of money paid in the n instalments. Then,

S30 = ₹24,000

Also,

S40 = ₹36,000

Subtracting (1) from (2), we get

$\left(2a+39d\right)-\left(2a+29d\right)=1800-1600\phantom{\rule{0ex}{0ex}}⇒10d=200\phantom{\rule{0ex}{0ex}}⇒d=20$

Putting d = 20 in (1), we get

$2a+29×20=1600\phantom{\rule{0ex}{0ex}}⇒2a+580=1600\phantom{\rule{0ex}{0ex}}⇒2a=1600-580=1020\phantom{\rule{0ex}{0ex}}⇒a=510$

Thus, the value of the first instalment is ₹510.

#### Question 48:

A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he delayed the work by 30 days?

It is given that the penalty for each succeeding day is ₹50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50.

Number of days in the delay of the work = 30

The amount of penalties are ₹200, ₹250, ₹300,... up to 30 terms.

∴ Total amount of money paid by the contractor as penalty, S30 = ₹200 + ₹250 + ₹300 + ... up to 30 terms

Here, a = ₹200, d = ₹50 and n = 30

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Hence, the contractor has to pay ₹27,750 as penalty.

#### Question 1:

Choose the correct answer in each of the following questions:

The common difference of the AP $\frac{1}{p},\frac{1-p}{p},\frac{1-2p}{p},...$ is                                                [CBSE 2013]

(a) p                                (b) −p                                (c) 1                                 (d) 1

The given AP is $\frac{1}{p},\frac{1-p}{p},\frac{1-2p}{p},...$ .

∴ Common difference, d = $\frac{1-p}{p}-\frac{1}{p}=\frac{1-p-1}{p}=\frac{-p}{p}=-1$

Hence, the correct answer is option C.

#### Question 2:

Choose the correct answer in each of the following questions:

The common difference of the AP $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$ is                                                [CBSE 2013]

(a) $\frac{1}{3}$                                (b) $\frac{-1}{3}$                                (c) b                                 (d) −b

The given AP is $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$ .

∴ Common difference, d = $\frac{1-3b}{3}-\frac{1}{3}=\frac{1-3b-1}{3}=\frac{-3b}{3}=-b$

Hence, the correct answer is option D.

#### Question 3:

Choose the correct answer in each of the following questions:

The next term of the AP is                                                            [CBSE 2014]

(a) $\sqrt{70}$                         (a) $\sqrt{84}$                          (a) $\sqrt{98}$                          (d) $\sqrt{112}$

The given terms of the AP can be written as i.e. .

∴ Next term = $4\sqrt{7}=\sqrt{16×7}=\sqrt{112}$

Hence, the correct answer is option D.

#### Question 4:

If 4, x1, x2, x3, 28 are in AP, then x3 = ?
(a) 19
(b) 23
(c) 22
(d) cannot be determined

(c) 22

Here, a = 4, l = 28 and n = 5
Then, ​T5 = 28
⇒​ a + (n - 1)d = 28
⇒​ 4 + (5 - 1)d= 28
⇒ 4d = 24
d = 6
Hence, x3 = 28 - 6 = 22

#### Question 5:

Choose the correct answer in each of the following questions:

If the nth term of the AP is (2n + 1) then the sum of its first three terms is

(a) 6n + 3                            (b) 15                            (c) 12                            (d) 21                      [CBSE 2012]

nth term of the AP, an = 2n + 1            (Given)

∴ First term, a1 = 2 × 1 + 1 = 2 + 1 = 3

Second term, a2 = 2 × 2 + 1 = 4 + 1 = 5

Third term, a3 = 2 × 3 + 1 = 6 + 1 = 7

∴ Sum of the first three terms = a1 + a2 + a3 = 3 + 5 + 7 = 15

Hence, the correct answer is option B.

#### Question 6:

Choose the correct answer in each of the following questions:

The sum of first n terms of an AP is (3n2 + 6n). The common difference of the AP is                   [CBSE 2014]

(a) 6                              (b) 9                              (c) 15                              (d) −3

Let Sn denotes the sum of first n terms of the AP.

So,
nth term of the AP, an = Sn − Sn − 1
$=\left(3{n}^{2}+6n\right)-\left(3{n}^{2}-3\right)\phantom{\rule{0ex}{0ex}}=6n+3$

Let d be the common difference of the AP.

Thus, the common difference of the AP is 6.

Hence, the correct answer is option A.

#### Question 7:

Choose the correct answer in each of the following questions:

The sum of first n terms of an AP is (5nn2). The nth term of the AP is

(a) (5 − 2n)                              (b) (6 − 2n)                              (c) (2n − 5)                              (d) (2n − 6)

Let Sn denotes the sum of first n terms of the AP.

nth term of the AP, an = Sn − Sn − 1
$=\left(5n-{n}^{2}\right)-\left(7n-{n}^{2}-6\right)\phantom{\rule{0ex}{0ex}}=6-2n$

Thus, the nth term of the AP is (6 − 2n).

Hence, the correct answer is option B.

#### Question 8:

Choose the correct answer in each of the following questions:

The sum of first n terms of an AP is (4n2 + 2n). The nth term of the AP is

(a) (6n − 2)                              (b) (7n − 3)                              (c) (8n − 2)                              (d) (8n + 2)                 [CBSE 2014]

Let Sn denotes the sum of first n terms of the AP.

nth term of the AP, an = Sn − Sn − 1
$=\left(4{n}^{2}+2n\right)-\left(4{n}^{2}-6n+2\right)\phantom{\rule{0ex}{0ex}}=8n-2$

Thus, the nth term of the AP is (8− 2).

Hence, the correct answer is option C.

#### Question 9:

Choose the correct answer in each of the following questions:

The 7th term of an AP is −1 and its 16th term is 17. The nth term of the AP is                                                [CBSE 2014]

(a) (3n + 8)                              (b) (4n − 7)                              (c) (15 − 2n)                              (d) (2n − 15)

Let a be the first term and d be the common difference of the AP. Then,

nth term of the AP, an = a + (− 1)d

Now,

Also,

Subtracting (1) from (2), we get

$\left(a+15d\right)-\left(a+6d\right)=17-\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒9d=18\phantom{\rule{0ex}{0ex}}⇒d=2$

Putting d = 2 in (1), we get

$a+6×2=-1\phantom{\rule{0ex}{0ex}}⇒a=-1-12=-13$

nth term of the AP, an = −13 + (− 1) × 2 = 2− 15

Hence, the correct answer is option D.

#### Question 10:

Choose the correct answer in each of the following questions:

The 5th term of an AP is −3 and its common difference is −4. The sum of its first 10 terms is                                     [CBSE 2011]

(a) 50                                  (b) −50                                  (c) 30                                  (d) −30

Let a be the first term of the AP.

Here, d = −4

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Thus, the sum of its first 10 terms is −50.

Hence, the correct answer is option B.

#### Question 11:

Choose the correct answer in each of the following questions:

The 5th term of an AP is 20 and the sum of its 7th and 11th terms is 64. The common difference of the AP is                  [CBSE 2015]

(a) 4                                  (b) 5                                  (c) 3                                  (d) 2

Let a be the first term and d be the common difference of the AP. Then,

Now,

${a}_{7}+{a}_{11}=64$                (Given)

From (1) and (2), we get

$20-4d+8d=32\phantom{\rule{0ex}{0ex}}⇒4d=32-20=12\phantom{\rule{0ex}{0ex}}⇒d=3$

Thus, the common difference of the AP is 3.

Hence, the correct answer is option C.

#### Question 12:

Choose the correct answer in each of the following questions:

The 13th term of an AP is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms is                          [CBSE 2015]

(a) 150                                  (b) 175                                  (c) 160                                  (d) 135

Let a be the first term and d be the common difference of the AP. Then,

Also,

Solving (1) and (2), we get

$a+3a=16\phantom{\rule{0ex}{0ex}}⇒4a=16\phantom{\rule{0ex}{0ex}}⇒a=4$

Putting a = 4 in (1), we get

$4d=3×4=12\phantom{\rule{0ex}{0ex}}⇒d=3$

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Thus, the sum of its first 10 terms is 175.

Hence, the correct answer is option B.

#### Question 13:

Choose the correct answer in each of the following questions:

An AP 5, 12, 19, ... has 50 terms. Its last term is                          [CBSE 2015]

(a) 343                                  (b) 353                                  (c) 348                                  (d) 362

The given AP is 5, 12, 19, ... .

Here, a = 5, d = 12 − 5 = 7 and n = 50

Since there are 50 terms in the AP, so the last term of the AP is a50.

Thus, the last term of the AP is 348.

Hence, the correct answer is option C.

#### Question 14:

Choose the correct answer in each of the following questions:

The sum of first 20 odd natural numbers is                             [CBSE 2012]

(a) 100                                  (b) 210                                  (c) 400                                  (d) 420

The first 20 odd natural numbers are 1, 3, 5, ..., 39.

These numbers are in AP.

Here, a = 1, l = 39 and n = 20

∴ Sum of first 20 odd natural numbers

Hence, the correct answer is option C.

#### Question 15:

Choose the correct answer in each of the following questions:

The sum of first 40 positive integers divisible by 6 is                                   [CBSE 2014]

(a) 2460                                  (b) 3640                                  (c) 4920                                  (d) 4860

The positive integers divisible by 6 are 6, 12, 18, ... .

This is an AP with a = 6 and d = 6.

Also, n = 40           (Given)

Using the formula, ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we get

Thus, the required sum is 4920.

Hence, the correct answer is option C.

#### Question 16:

Choose the correct answer in each of the following questions:

How many two-digit numbers are divisible by 3?                                   [CBSE 2012]

(a) 25                                  (b) 30                                  (c) 32                                  (d) 36

The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.

Clearly, these number are in AP.

Here, a = 12 and d = 15 − 12 = 3

Let this AP contains n terms. Then,

$⇒n=30$

Thus, there are 30 two-digit numbers divisible by 3.

Hence, the correct answer is option is B.

#### Question 17:

Choose the correct answer in each of the following questions:

How many three-digit numbers are divisible by 9?                                   [CBSE 2013]

(a) 86                                  (b) 90                                  (c) 96                                  (d) 100

The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.

Clearly, these numbers are in AP.

Here, a = 108 and d = 117 − 108 = 9

Let this AP contains n terms. Then,

Thus, there are 100 three-digit numbers divisible by 9.

Hence, the correct answer is option D.

#### Question 18:

Choose the correct answer in each of the following questions:

What is the common difference of an AP in which ${a}_{18}-{a}_{14}=32$?

(a) 8                                  (b) −8                                  (c) 4                                  (d) −4

Let a be the first term and d be the common difference of the AP. Then,

Thus, the common difference of the AP is 8.

Hence, the correct answer is option is A.

#### Question 19:

Choose the correct answer in each of the following questions:

If an denotes the nth term of the AP 3, 8, 13, 18, ... then what is the value of $\left({a}_{30}-{a}_{20}\right)$?

(a) 40                                  (b) 36                                  (c) 50                                  (d) 56

The given AP is 3, 8, 13, 18, ... .

Here, a = 3 and d = 8 − 3 = 5

Thus, the required value is 50.

Hence, the correct answer is option C.

#### Question 20:

Choose the correct answer in each of the following questions:

Which term of the AP 72, 63, 54, ... is 0?

(a) 8th                                  (b) 9th                                  (c) 10th                                  (d) 11th

The given AP is 72, 63, 54, ... .

Here, a = 72 and d = 63 − 72 = −9

Suppose nth term of the given AP is 0. Then,

Thus, the 9th term of the given AP is 0.

Hence, the correct answer is option B.

#### Question 21:

Choose the correct answer in each of the following questions:

Which term of the AP 25, 20, 15, ... is the first negative term?

(a) 10th                                  (b) 9th                                  (c) 8th                                  (d) 7th

The given AP is 25, 20, 15, ... .

Here, a = 25 and d = 20 − 25 = −5

Let the nth term of the given AP be the first negative term. Then,

Thus, the 7th term is the first negative term of the given AP.

Hence, the correct answer is option D.

#### Question 22:

Which term of the AP 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th

(b) 10th

Here, a = 21 and d = (42 - 21) = 21
Let 210 be the nth term of the given AP.
Then ​Tn = 210
⇒​ a + (n - 1)d = 210
⇒​ 21 + (n - 1) ⨯ 21= 210
⇒ 21n = 210
n = 10
Hence, 210 is the 10th term of the AP.

#### Question 23:

Choose the correct answer in each of the following questions:

What is 20th term from the end of the AP 3, 8, 13, ..., 253?

(a) 163                                  (b) 158                                  (c) 153                                  (d) 148

The given AP is 3, 8, 13, ..., 253.

Let us re-write the given AP in reverse order i.e. 253, 248, ..., 13, 8, 3.

Now, the 20th term from the end of the given AP is equal to the 20th term from beginning of the AP 253, 248, ..., 13, 8, 3.

Consider the AP 253, 248, ..., 13, 8, 3.

Here, a = 253 and d = 248 − 253 = −5

∴ 20th term of this AP

= 253 + (20 − 1) × (−5)

= 253 − 95

= 158

Thus, the 20th term from the end of the given AP is 158.

Hence, the correct answer is option B.

#### Question 24:

(5 + 13 + 21 + ... + 181) = ?
(a) 2476
(b) 2337
(c) 2219
(d) 2139

(d) 2139

Here, = 5, d = (13-5) = 8 and l = 181
Let the number of terms be n.
Then Tn = 181

a + (n-1d = 181
⇒ 5 + ( n -1​) ⨯​ 8 = 181
⇒  8n = 184
⇒ n = 23

∴ Required sum =

Hence, the required sum is 2139.

#### Question 25:

The sum of first 16 terms of the AP 10, 6, 2, ..., is
(a) 320
(b) −320
(c) −352
(d) −400

(b) - 320

Here, a = 10, = (6 - 10) = -4 and n = 16
Using the formula, , we get:

Hence, the sum of the first 16 terms of the given AP is -320.

#### Question 26:

How many terms of the AP 3, 7, 11, 15, ... will make the sum 406?
(a) 10
(b) 12
(c) 14
(d) 20

(c) 14
Here, = 3 and d = (7-3) = 4
Let the sum of n terms be 406 .
Then,
we have:

Hence, 14 terms will make the sum 406.

#### Question 27:

The 2nd term of an AP is 13 and its 5th term is 25. What is its 17th term.
(a) 69
(b) 73
(c) 77
(d) 81

(b) 73

T2 = a + d  = 13              ...(i)
T5 = a + 4d  = 25            ...(ii)
On subtracting (i) from (ii), we get:
⇒ 3d = 12
⇒ d = 4
On putting the value of d in (i), we get:
⇒ a + 4= 13
⇒ a = 9

Now, T17 = a +16d = 9 + 16 ⨯ 4 = 73
Hence, the 17th term is 73.

#### Question 28:

The 17th term of an AP exceeds its 10th term by 21. The common difference of the AP is
(a) 3
(b) 2
(c) −3
(d) −2

( a) 3
T10 = a + 9d
T17 = a + 16d

Also, a + 16d = 21 + T10
⇒ a + 16d = 21 + a + 9d
⇒ 7d = 21
⇒ d = 3
Hence, the common difference of the AP is 3.

#### Question 29:

The 8th term of an AP is 17 and its 14th term is 29. The common difference of the AP is
(a) 3
(b) 2
(c) 5
(d) −2

( b) 2
T8 = a + 7d = 17     ...(i)
T14 = a + 13d = 29    ...(ii)

On subtracting (i) from (ii), we get:
⇒ 6d = 12
d = 2
Hence, the common difference is 2.

#### Question 30:

The 7th term of an AP is 4 and its common difference is −4. What is its first term?
(a) 16
(b) 20
(c) 24
(d) 28