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

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

The given A.P. is 1, 4, 7, 10, ……….

For a given A.P.,

Number of terms, n = 60

First term, a = 7

Last term, l = 125

⇒ t_{60}
= 125

⇒ a + 59d = 125

⇒ 7 + 59d = 125

⇒ 59d = 118

⇒ d = 2

Hence,
t_{31} = a + 30d = 7 + 30(2) = 7 + 60 = 67

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

t_{4}
+ t_{8} = 24 (given)

⇒ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 ….(i)

And,

t_{6}
+ t_{10} = 34 (given)

⇒ (a + 5d) + (a + 9d) = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17 ….(ii)

Subtracting (i) from (ii), we get

2d = 5

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

Now,
t_{3} = 5 (given)

⇒ a + 2d = 5 ….(i)

And,

t_{7}
= 9 (given)

⇒ a + 6d = 9 ….(ii)

Subtracting (i) from (ii), we get

4d = 4

⇒ d = 1

⇒ a + 2(1) = 5

⇒ a = 3

Hence,
17^{th} term = t_{17} = a + 16d = 3 + 16(1) = 19

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

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

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

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

n^{th} term of an A.P. is given by

t_{n} = a + (n - 1)d

⇒ a_{5} = 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.

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

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

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

First instalment = a = Rs. 1000

Increase in intalment every month = d= Rs. 100

30th instalment = t_{30}

= a + 29d

= 1000 + 29 × 100

= 1000 + 2900

= Rs. 3900

Now, amount paid in 30 instalments = S_{30}

= 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)

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

Now,
t_{6} = 16 (given)

⇒ a + 5d = 16 ….(i)

And,

t_{14}
= 32 (given)

⇒ a + 13d = 32 ….(ii)

Subtracting (i) from (ii), we get

8d = 16

⇒ d = 2

⇒ a + 5(2) = 16

⇒ a = 6

Hence,
36^{th} term = t_{36} = a + 35d = 6 + 35(2) = 76

For a given A.P.,

Number of terms, n = 50

3^{rd}
term, t_{3} = 12

⇒ a + 2d = 12 ….(i)

Last term, l = 106

⇒ t_{50}
= 106

⇒ a + 49d = 106 ….(ii)

Subtracting (i) from (ii), we get

47d = 94

⇒ d = 2

⇒ a + 2(2) = 12

⇒ a = 8

Hence,
t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64

Here,

First term, a = 4

Common difference, d = 6 - 4 = 2

n = 10

Here,

First term, a = 3

Last term, l = 57

n = 20

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 - n^{2}

⇒ n^{2}
- 13n + 30 = 0

⇒ n^{2}
- 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.

t_{n} = 8 - 5n

Replacing n by (n + 1), we get

t_{n+1}
= 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n

Now,

t_{n+1}
- t_{n} = (3 - 5n) - (8 - 5n) = -5

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

Hence, the given sequence is an A.P.

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 (n^{th} term) of an A.P. is given by

t_{n} = a + (n - 1)d

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

= 3 - 2n + 2

= 5 - 2n

Hence, 23^{rd} term = t_{23} = 5 - 2(23) = 5 - 46 = -41

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 n^{th} term of the given
A.P. be 78.

⇒ 78 = 3 + (n - 1)(5)

⇒ 75 = 5n - 5

⇒ 5n = 80

⇒ n = 16

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

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

t_{n} = 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.

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

Given,

S_{14}
= 1050

⇒ 7[2a + 13d] = 1050

⇒ 2a + 13d = 150

⇒ a + 6.5d = 75 ….(i)

And,
t_{14} = 140

⇒ a + 13d = 140 ….(ii)

Subtracting (i) from (ii), we get

6.5d = 65

⇒ d = 10

⇒ a + 13(10) = 140

⇒ a = 10

Thus,
20^{th} term = t_{20} = 10 + 19d = 10 + 19(10) = 200

n^{th}
term of an A.P. is given by t_{n}_{}=
a + (n - 1) d.

⇒ t_{25 } = a + (25 - 1)d = a + 24d and

t_{9} = a + (9 - 1)d = a + 8d

According to the condition in the question, we get

t_{25 }=
t_{9} + 16

⇒ a + 24d = a + 8d + 16

⇒ 16d = 16

⇒ d = 1

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 × m^{th}^{}term

Hence proved.

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

a = 58 and d = 2

t_{n} = a + (n - 1)d

⇒ t_{n}_{} = 58 + (n - 1)2 …. (i)

In the first A.P. -2, 5, 12, ….

a = -2 and d = 7

t_{n} = a + (n - 1)d

⇒ t_{n}_{}=
-2 + (n - 1)7 …. (ii)

Given that the n^{th} term of first A.P is equal
to the n^{th} 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

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

Let a_{n} be the
first negative term.

⇒ a_{n} < 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 28^{th} term is the first negative
term of the given A.P.

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 t_{n}
= 994.

We know that n^{th} term of A.P is given by

t_{n} = 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.

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.

We have 2n^{2} - 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 n^{th} term of an A.P can't be 2n^{2}
- 7.

Here a = 14 , d =
7 and t_{n} = 168

t_{n} = 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.

Here a = 20 and S_{7}
= 2100

We know that,

To find: t_{31}
=?

t_{n} = a + (n - 1)d

Therefore the 31^{st} term of the given A.P. is
2820.** **

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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