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# Class 10 SELINA Solutions Maths Chapter 10 - Arithmetic Progression

## Arithmetic Progression Exercise 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

## Arithmetic Progression Exercise 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 ## Arithmetic Progression Exercise 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 ## Arithmetic Progression Exercise 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 ## Arithmetic Progression Exercise 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 instalment every month = d= Rs. 100

30th instalment = t30

= a + 29d

= 1000 + 29 × 100

= 1000 + 2900

= Rs. 3900

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

= Rs. 73, 500

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

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

= Rs. 44,500

## Arithmetic Progression Exercise 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.