# SELINA Solutions for Class 10 Maths Chapter 10 - Arithmetic Progression

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## Chapter 10 - Arithmetic Progression Ex. 10(A)

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11

The given A.P. is 1, 4, 7, 10, ………. Solution 12 Solution 13 Solution 14 Solution 15 Solution 16 Solution 17 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

## Chapter 10 - Arithmetic Progression Ex. 10(B)

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 Solution 15 Solution 16 ## Chapter 10 - Arithmetic Progression Ex. 10(C)

Solution 1 Solution 2 Solution 3 Solution 4(i) Solution 4(ii) Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 ## Chapter 10 - Arithmetic Progression Ex. 10(D)

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5

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 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 ## Chapter 10 - Arithmetic Progression Ex. 10(E)

Solution 1 Solution 2 Solution 3 Solution 4

Since the production increases uniformly by a fixed number every year, he sequence formed by the production in different years is an A.P.

Let the production in the first year = a

Common difference = Number of units by which the production increases every year = d Solution 5

Total amount of loan = Rs. 1,18,00

First instalment = a = Rs. 1000

Increase in intalment every month = d= Rs. 100

30th instalment = t30

= a + 29d

= 1000 + 29 × 100

= 1000 + 2900

= Rs. 3900

Now, amount paid in 30 instalments = S30 = 15 × 4900

= Rs. 73, 500

∴ Amount of loan to be paid after the 30th instalments

= Rs. (1,18,000 - 73500)

= Rs. 44,500

## Chapter 10 - Arithmetic Progression Ex. 10(F)

Solution 1

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 2 Solution 3

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 4 Solution 5

Here,

First term, a = 4

Common difference, d = 6 - 4 = 2

n = 10 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 3, 8, 13, …..

Now,

8 - 3 = 13 - 8 = 5

Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.

Let the nth term of the given A.P. be 78.

78 = 3 + (n - 1)(5)

75 = 5n - 5

5n = 80

n = 16

Thus, the 16th term of the given sequence is 78.

Solution 11

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 12 Solution 13 Solution 14 Solution 15

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 16

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 17

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 18

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 19

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 20

The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.

This is an A.P. in which a = 105, d = 7 and tn = 994.

We know that nth term of A.P is given by

tn = a + (n - 1)d.

994 = 105 + (n - 1)7

889 = 7n - 7

896 = 7n

n = 128

There are 128 three digit numbers which are divisible by 7.

Solution 21

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 22

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 23

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 24

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 25

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.

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