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#### Page No 260:

(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 = $5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$

#### Page No 261:

(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

#### Page No 261:

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

(ii) The given AP is
First term = 5
Common difference

#### Page No 261:

And the numbers are:

#### Page No 261:

$\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}16,9,2,-5\phantom{\rule{0ex}{0ex}}{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}a=16\phantom{\rule{0ex}{0ex}}d=9-16=-7\phantom{\rule{0ex}{0ex}}{a}_{n}=16+\left(n-1\right)\left(-7\right)\phantom{\rule{0ex}{0ex}}{a}_{n}=16-7n+7\phantom{\rule{0ex}{0ex}}{a}_{n}=23-7n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 261:

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

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

In the given problem, let us first find the 41st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. (d= 6

Now, as we know,

So, for 41st term (n = 41),

Let us take the term which is 72 more than the 41st term as an. So,

Also,

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41st term.

#### Page No 261:

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.

#### Page No 261:

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,...

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 261:

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.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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,...

#### Page No 262:

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.

#### Page No 262:

Thus, (1) = (2)
Hence, 33rd term is three times its 15th term.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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).

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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.

#### Page No 262:

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,...

#### Page No 262:

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.

#### Page No 262:

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).

#### Page No 262:

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.

#### Page No 262:

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, ... .

#### Page No 262:

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).

#### Page No 263:

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) .

#### Page No 263:

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.

#### Page No 263:

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.

#### Page No 263:

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.

#### Page No 263:

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.

#### Page No 263:

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.

#### Page No 263:

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.

#### Page No 263:

Numbers between 200 and 500 divisible by 8 are 208, 216, ..., 496.
This forms an AP 208, 216, ..., 496.
So, first term (a) = 208
Common difference (d) = 8
${a}_{n}=a+\left(n-1\right)d=496\phantom{\rule{0ex}{0ex}}⇒208+\left(n-1\right)8=496\phantom{\rule{0ex}{0ex}}⇒\left(n-1\right)8=288\phantom{\rule{0ex}{0ex}}⇒n-1=36\phantom{\rule{0ex}{0ex}}⇒n=37$
Thus, there are 37 integers between 200 and 500 which are divisible by 8.

#### Page No 267:

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.

#### Page No 267:

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.

#### Page No 267:

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.

#### Page No 267:

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

#### Page No 267:

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.

#### Page No 267:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 15.

a − d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5

Their product is 105.

Hence, the three numbers are 3, 5, 7 or 7, 5, 3.

#### Page No 267:

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).

#### Page No 267:

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).

#### Page No 267:

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).

#### Page No 268:

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

#### Page No 268:

${a}^{2}+9{d}^{2}-6ad+{a}^{2}+{d}^{2}-2ad+{a}^{2}+{d}^{2}+2ad+{a}^{2}+9{d}^{2}+6ad=216\phantom{\rule{0ex}{0ex}}⇒4{a}^{2}+20{d}^{2}=216\phantom{\rule{0ex}{0ex}}⇒4{\left(7\right)}^{2}+20{d}^{2}=216\phantom{\rule{0ex}{0ex}}⇒d=±1\phantom{\rule{0ex}{0ex}}$

#### Page No 268:

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.

#### Page No 268:

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.

#### Page No 268:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6

The product of the first and third term is 5 times the common difference.

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.

#### Page No 285:

(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

#### Page No 285:

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

Using formula of sum

Using formula of sum

#### Page No 285:

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

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

#### Page No 285:

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.

#### Page No 285:

Given:

Hence,
(i) nth term is (6n – 7)
(ii) first term is –1
(iii) common difference is 6

#### Page No 285:

(i)

(ii) 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.

#### Page No 285:

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

And,

Subtracting (2) from (1), we get
$\frac{1}{n}-\frac{1}{m}=\left(m-n\right)d\phantom{\rule{0ex}{0ex}}⇒\frac{m-n}{mn}=\left(m-n\right)d\phantom{\rule{0ex}{0ex}}⇒d=\frac{1}{mn}$
Putting $d=\frac{1}{mn}$ in (1), we get

∴ Sum of mn terms,

#### Page No 285:

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.

#### Page No 285:

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.

#### Page No 285:

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.

#### Page No 285:

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.

#### Page No 285:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

Let the given series be X = $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+...$
$=\left[4+4+4+...\right]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...\right]\phantom{\rule{0ex}{0ex}}=4\left[1+1+1+...\right]-\frac{1}{n}\left[1+2+3+...\right]\phantom{\rule{0ex}{0ex}}={S}_{1}-{S}_{2}$

Hence, the sum of n terms of the series is $\frac{7n-1}{2}$.

#### Page No 286:

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, ... .

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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.

#### Page No 286:

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 $\frac{8}{3}$.

#### Page No 286:

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

#### Page No 286:

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.

#### Page No 286:

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, ... .

#### Page No 286:

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 =

#### Page No 286:

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.

#### Page No 287:

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, ... .

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

(i) 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.

(ii) 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.

#### Page No 287:

Let the first term of the first AP be a
And common difference be d

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 287:

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.

#### Page No 288:

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.

#### Page No 288:

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.

#### Page No 288:

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.

#### Page No 288:

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.

#### Page No 288:

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.

#### Page No 288:

Saving of the child on the first day = Rs 5
Saving on the second day = Rs 5 + Rs 5 = Rs 10
Saving on the third day = Rs 5 + 2 × Rs 5 = Rs 15 and so on
The saving of the child on different days are Rs 5, Rs 10, Rs 15, ....
Since the savings of the child for each succeeding day is Rs 5 more than for the preceeding day, therefore the savings for different days forms an AP with first term a = Rs 5 and common difference d = Rs 5.
Suppose the number of days she continued to put the five-rupees coin in the piggy bank be n.
It is given that the total number of five-rupees coins in the piggy bank is 190.
So, the total sum of money saved by the child in n days = 190 × 5 = Rs 950

Since the number of days cannot be negative, so n = 19.
So, the number of days she continued to put the five-rupees coin in the piggy bank is 19.
Also,
Total sum of money saved by her = Rs 950

#### Page No 292:

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.

#### Page No 292:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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

Here, a = 2 and d = 7 − 2 = 5

Hence, the required value is 50.

#### Page No 293:

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.

#### Page No 293:

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.

:

#### Page No 293:

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.

#### Page No 293:

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

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

#### Page No 293:

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).

#### Page No 293:

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

#### Page No 293:

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 = $\frac{7}{5}$

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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.

#### Page No 293:

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, ... .

#### Page No 293:

$n=14\phantom{\rule{0ex}{0ex}}{s}_{n}=\frac{n}{2}\left[a+l\right]\phantom{\rule{0ex}{0ex}}287=\frac{14}{2}\left[1+x\right]\phantom{\rule{0ex}{0ex}}⇒x=40$

#### Page No 295:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

(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

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 296:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

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.

#### Page No 297:

(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.

#### Page No 297:

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.

#### Page No 297:

(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.

#### Page No 297:

(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.

#### Page No 297:

(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.

#### Page No 297:

(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.

#### Page No 297:

( 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.

#### Page No 297:

( 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.