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# Class 10 SELINA Solutions Maths Chapter 10 - Arithmetic Progression

## Arithmetic Progression Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (iv) All are A.P.

A + B = (2 + 3), (4 + 6), (6 + 9), (8 + 12), …… = 5, 10, 15, 20, ….. is an A.P.

A – C = (2 – 0), (4 – 4), (6 – 8), (8 – 12), …. = 2, 0, –2, –4, …. is an A.P.

C – B = (0 – 3), (4 – 6), (8 – 9), (12 – 12), ….. = –3, –2, –1, 0, …. is an A.P.

B – A = (3 – 2), (6 – 4), (9 – 6), (12 – 8), ….. = 1, 2, 3, 4, …. is an A.P.

### Solution 1(b)

Correct option: (iv) 5

2k – 7, k + 5 and 3k + 2 are in A.P.

(k + 5) – (2k – 7) = (3k + 2) – (k + 5)

12 – k = 2k – 3

3k = 15

k = 5

### Solution 1(c)

Correct option: (ii) 5

l = a + (n – 1)d

36 = –36 + (n – 1)18

72 = (n – 1)18

n – 1 = 4

n = 5

### Solution 1(d)

Correct option: (iii) Yes, 97

12, 52, 72, 73 …. = 1, 25, 49, 73 ….

This forms an A.P. with first term 1 and common difference 24.

Therefore, next term = 73 + 24 = 97

### Solution 1(e)

Correct option: (ii) 110

First 10 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

First term = 2

Common difference = 2

Options (i) and (iv) are same in textbook.

### Solution 1(f)

Correct option: (i)

The statement given in assertion is correct.

Hence, assertion A is true.

The difference between the consecutive terms is same, i.e. .

Hence, reason (R) is true.

### Solution 2

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Now, t6 = 16 (given)

a + 5d = 16 ….(i)

And,

t14 = 32 (given)

a + 13d = 32 ….(ii)

Subtracting (i) from (ii), we get

8d = 16

d = 2

a + 5(2) = 16

a = 6

Hence, 36th term = t36 = a + 35d = 6 + 35(2) = 76

### Solution 4

For a given A.P.,

Number of terms, n = 50

3rd term, t3 = 12

a + 2d = 12 ….(i)

Last term, l = 106

t50 = 106

a + 49d = 106 ….(ii)

Subtracting (i) from (ii), we get

47d = 94

d = 2

a + 2(2) = 12

a = 8

Hence, t29 = a + 28d = 8 + 28(2) = 8 + 56 = 64

### Solution 6

Here,

First term, a = 3

Last term, l = 57

n = 20

### Solution 7

Here, we find that

15 - 18 = 12 - 15 = -3

Thus, the given series is an A.P. with first term 18 and common difference -3.

Let the number of term to be added be 'n'.

90 = n[36 - 3n + 3]

90 = n[39 - 3n]

90 = 3n[13 - n]

30 = 13n - n2

n2 - 13n + 30 = 0

n2 - 10n - 3n + 30 = 0

n(n - 10) - 3(n - 10) = 0

(n - 10)(n - 3) = 0

n - 10 = 0 or n - 3 = 0

n = 10 or n = 3

Thus, required number of term to be added is 3 or 10.

### Solution 8

tn = 8 - 5n

Replacing n by (n + 1), we get

tn+1 = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n

Now,

tn+1 - tn = (3 - 5n) - (8 - 5n) = -5

Since, (tn+1 - tn) is independent of n and is therefore a constant.

Hence, the given sequence is an A.P.

### Solution 9

The given sequence is 1, -1, -3, …..

Now,

1 - 3 = -1 - 1 = -3 - (-1) = -2

Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.

The general term (nth term) of an A.P. is given by

tn = a + (n - 1)d

= 3 + (n - 1)(-2)

= 3 - 2n + 2

= 5 - 2n

Hence, 23rd term = t23 = 5 - 2(23) = 5 - 46 = -41

### Solution 10

The given sequence is 11, 8, 5, 2, …..

Now,

8 - 11 = 5 - 8 = 2 - 5 = -3

Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.

The general term of an A.P. is given by

tn = a + (n - 1)d

-150 = 11 + (n - 1)(-5)

-161 = -5n + 5

5n = 166

The number of terms cannot be a fraction.

So, clearly, -150 is not a term of the given sequence.

### Solution 13

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Given,

S14 = 1050

7[2a + 13d] = 1050

2a + 13d = 150

a + 6.5d = 75 ….(i)

And, t14 = 140

a + 13d = 140 ….(ii)

Subtracting (i) from (ii), we get

6.5d = 65

d = 10

a + 13(10) = 140

a = 10

Thus, 20th term = t20 = 10 + 19d = 10 + 19(10) = 200

### Solution 14

nth term of an A.P. is given by tn= a + (n - 1) d.

t25  = a + (25 - 1)d = a + 24d and

t9 = a + (9 - 1)d = a + 8d

According to the condition in the question, we get

t25 = t9 + 16

a + 24d = a + 8d + 16

16d = 16

d = 1

### Solution 15

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

(m + n)th term = a + (m + n - 1)d …. (i) and

(m - n)th term = a + (m - n - 1)d …. (ii)

From (i) + (ii), we get

(m + n)th term + (m - n)th term

= a + (m + n - 1)d + a + (m - n - 1)d

= a + md + nd - d + a + md - nd - d

= 2a + 2md - 2d

= 2a + (m - 1)2d

= 2[ a + (m - 1)d]

= 2 × mthterm

Hence proved.

### Solution 16

In the first A.P. 58, 60, 62,....

a = 58 and d = 2

tn = a + (n - 1)d

tn = 58 + (n - 1)2 …. (i)

In the first A.P. -2, 5, 12, ….

a = -2 and d = 7

tn = a + (n - 1)d

tn= -2 + (n - 1)7 …. (ii)

Given that the nth term of first A.P is equal to the nth term of the second A.P.

58 + (n - 1)2 = -2 + (n - 1)7 … from (i) and (ii)

58 + 2n - 2 = -2 + 7n - 7

65 = 5n

n = 15

### Solution 17

Here a = 105 and d = 101 - 105 = -4

Let an be the first negative term.

an < 0

a + (n - 1)d < 0

105 + (n - 1)(-4)<0

105 - 4n + 4 <0

109 - 4n < 0

109 <4n

27.25 < n

The value of n = 28.

Therefore 28th term is the first negative term of the given A.P.

### Solution 18

Let the three parts of 216 in A.P be (a - d), a, (a + d).

a - d + a + a + d = 216

3a = 216

a = 72

Given that the product of the two smaller parts is 5040.

a(a - d ) = 5040

72(72 - d) = 5040

72 - d = 70

d = 2

a - d = 72 - 2 = 70, a = 72 and a + d = 72 + 2 = 74

Therefore the three parts of 216 are 70, 72 and 74.

### Solution 19

We have 2n2 - 7,

Substitute n = 1, 2, 3, … , we get

2(1)2 - 7, 2(2)2 - 7, 2(3)2 - 7, 2(4)2 - 7, ….

-5, 1, 11, ….

Difference between the first and second term = 1 - (-5) = 6

And Difference between the second and third term = 11 - 1 = 10

Here, the common difference is not same.

Therefore the nth term of an A.P can't be 2n2 - 7.

### Solution 20

Here a = 14 , d = 7 and tn = 168

tn = a + (n - 1)d

168 = 14 + (n - 1)7

154 = 7n - 7

154 = 7n - 7

161 = 7n

n = 23

We know that,

Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.

### Solution 21

Here a = 20 and S7 = 2100

We know that,

To find: t31 =?

tn = a + (n - 1)d

Therefore the 31st term of the given A.P. is 2820.

### Solution 22

First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.

58, …., -8, -10, -12.

Here a = 58 , d = -2

Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.

## Arithmetic Progression Exercise Ex. 10(A)

### Solution 11

The given A.P. is 1, 4, 7, 10, ……….

### Solution 18

For a given A.P.,

Number of terms, n = 60

First term, a = 7

Last term, l = 125

t60 = 125

a + 59d = 125

7 + 59d = 125

59d = 118

d = 2

Hence, t31 = a + 30d = 7 + 30(2) = 7 + 60 = 67

### Solution 19

Let 'a' be the first term and 'd' be the common difference of the given A.P.

t4 + t8 = 24 (given)

(a + 3d) + (a + 7d) = 24

2a + 10d = 24

a + 5d = 12 ….(i)

And,

t6 + t10 = 34 (given)

(a + 5d) + (a + 9d) = 34

2a + 14d = 34

a + 7d = 17 ….(ii)

Subtracting (i) from (ii), we get

2d = 5

### Solution 20

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Now, t3 = 5 (given)

a + 2d = 5 ….(i)

And,

t7 = 9 (given)

a + 6d = 9 ….(ii)

Subtracting (i) from (ii), we get

4d = 4

d = 1

a + 2(1) = 5

a = 3

Hence, 17th term = t17 = a + 16d = 3 + 16(1) = 19

### Solution 1(a)

Correct option: (ii) 8, 3, –2, –7, …..

First term = a = 8

Common difference = d = –5

Then, the A.P. is 8, 8 + (–5), 8 + 2(–5), 8 + 3(–5), ……

That is, 8, 3, –2, –7, …..

### Solution 1(b)

Correct option: (ii) yes

The given sequence –8, –8, –8, –8, ….. is an A.P. with the first term –8 and common difference zero.

### Solution 1(c)

Correct option: (iv) –39

Given A.P. is 3, 0, –3, –6, …..

First term, a = 3

Common difference, d = 0 – 3 = –3

15th term, t15 = a + 14d

= 3 + 14(–3)

= 3 – 42

= –39

### Solution 1(d)

Correct option: (ii) 2

t24 – t19 = 10

(a + 23d) – (a + 18d) = 10

5d = 10

d = 2

### Solution 1(e)

Correct option: (ii) 16

Given A.P. is 8, 13, 18, …..

First term, a = 8

Common difference, d = 13 – 8 = 5

nth term, tn = 83

a + (n – 1)d = 83

8 + (n – 1)5 = 83

(n – 1)5 = 75

n – 1 = 15

n = 16

## Arithmetic Progression Exercise Ex. 10(B)

### Solution 1(a)

Correct option: (iii) 8

Common difference of two A.P.,s = d

Let the first and nth terms of the first A.P. be a and tn, respectively.

Let the first and nth terms of the second A.P. be A and Tn, respectively.

Then, t25 – T25 = 8

a + 24d – A – 24d = 8

a – A = 8

Now, t50 – T50 = a + 49d – A – 49d = a – A = 8

### Solution 1(b)

Correct option: (i) 0

Let the first term and the common difference of an A.P. be a and d, respectively.

Then, 10 × t10 = 20 × (t20)

10 × (a + 9d) = 20 × (a + 19d)

a + 9d = 2(a + 19d)

a + 9d = 2a + 38d

a + 29d = 0

t30 = 0

### Solution 1(c)

Correct option: (iv) 7

tn = 7n – 5

Then, t1 = 7(1) – 5 = 2 and t2 = 7(2) – 5 = 9

Therefore, common difference = t2 – t1 = 9 – 2 = 7

### Solution 1(d)

Correct option: (iv) 3

t40 – t16 = 72

(a + 39d) – (a + 15d) = 72

24d = 72

d = 3

### Solution 1(e)

Correct option: (i) 17

First term = a = 6

Common difference = d = 11 – 6 = 5

tn = 106

a + (n – 1)d = 106

6 + (n – 1)(5) = 106

(n – 1)(5) = 100

n – 1 = 20

n – 1 – 3 = 17

n – 4 = 17

## Arithmetic Progression Exercise Ex. 10(C)

### Solution 1(a)

Correct option: (ii) 1640

Total terms = 41

Then, middle term = 21st term = 40

That is, a + 20d = 40

### Solution 1(b)

Correct option: (iv) 4905

Two digit numbers are 10, 11, 12, ……99

This forms an A.P. with first term, a = 10, last term, l = 99 and common difference, d = 1

Hence, n = 100 – 10 = 90

### Solution 1(c)

Correct option: (i) 650

First term, a = 4

Common difference, d = 7 – 4 = 3

n = 20

### Solution 1(d)

Correct option: (ii) 2620

First term, a = 7

Common difference, d = 10 – 7 = 3

n = 40

### Solution 1(e)

Correct option: (iii) 26

tn = 6n + 4

Then, t1 = 6(1) + 4 = 10 and t2 = 6(2) + 4 = 16

Sum of first two terms = t1 + t2 = 10 + 16 = 26

## Arithmetic Progression Exercise Ex. 10(D)

### Solution 6

We know that,

Sum of n terms of an A.P =

Let the first term be 2x and the last term be 3x.

Sum of 5 terms of an A.P =

First term = 2x =2 × 1 = 2 and the last term = 3x = 3 × 1 = 3

nth term of an A.P. is given by

tn = a + (n - 1)d

a5 = 2 + (5 - 1)d

3 = 2 + 4d

1 = 4d

d =  = 0.25

Therefore the five numbers in an A.P are 2, 2.25, 2.50, 2.75 and 3.

### Solution 1(a)

Correct option: (iii) 2

t1 = k + 2, t2 = 2k + 7 and t3 = 4k + 12

t2 – t1 = t3 – t2

(2k + 7) – (k + 2) = (4k + 12) – (2k + 7)

k + 5 = 2k + 5

k = 0

Therefore, first term = 0 + 2 = 2

### Solution 1(b)

Correct option: (iii) 9

Sn = 3n2

S1 = 3(1)2 = 3

S2 = 3(2)2 = 12

Second term = S2 – S1= 12 – 3 = 9

### Solution 1(c)

Correct option: (ii) 5 × 7, 7 × 7 and 9 × 7

5, 7 and 9 are in A.P.

If each term of an A.P. is multiplied by a non-zero fixed number, the resulting sequence is an A.P.