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

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

Question 1

Which of the following sequences are in arithmetic progression?

(i) 2, 6, 10, 14,

(ii) 15, 12, 9, 6,

(iii) 5, 9, 12, 18,

(iv) Solution 1 Question 2

The nth term of sequence is (2n - 3), find its fifteenth term.

Solution 2 Question 3

If the pth term of an A.P. is (2p + 3), find the A.P.

Solution 3 Question 4

Find the 24th term of the sequence:

12, 10, 8, 6,……

Solution 4 Question 5

Find the 30th term of the sequence: Solution 5 Question 6 Solution 6 Question 7 Solution 7 Question 8

Is 402 a term of the sequence :

8, 13, 18, 23,………….?

Solution 8 Question 9 Solution 9 Question 10

How many terms are there in the series :

(i) 4, 7, 10, 13, …………, 148?

(ii) 0.5, 0.53, 0.56, ……………, 1.1? Solution 10 Question 11

Which term of the A.P. 1, 4, 7, 10, ………. is 52?

Solution 11

The given A.P. is 1, 4, 7, 10, ………. Question 12

If 5th and 6th terms of an A.P are respectively 6 and 5. Find the 11th term of the A.P

Solution 12 Question 13

If tn represents nth term of an A.P, t2 + t5 - t3 = 10 and t2 + t9 = 17, find its first term and its common difference.

Solution 13 Question 14

Find the 10th term from the end of the A.P. 4, 9, 14,…….., 254

Solution 14 Question 15

Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.

Solution 15 Question 16

Find the 31st term of an A.P whose 10th term is 38 and 16th term is 74.

Solution 16 Question 17

Which term of the series :

21, 18, 15, …………. is - 81?

Can any term of this series be zero? If yes find the number of term.

Solution 17 Question 18

An A.P. consists of 60 terms, If the first and the last terms be 7 and 125 respectively, find the 31st term.

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

Question 19

The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the 6th and the 10th terms of the same A.P. is 34. Find the first three terms of the A.P.

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 Question 20

If the third term of an A.P. is 5 and the seventh term is 9, find the 17th term.

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 Exercise Ex. 10(B)

Question 1

In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.

Solution 1 Question 2

How many two-digit numbers are divisible by 3?

Solution 2 Question 3

Which term of A.P. 5, 15, 25 ………… will be 130 more than its 31st term?

Solution 3 Question 4

Find the value of p, if x, 2x + p and 3x + 6 are in A.P

Solution 4 Question 5

If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?

Solution 5 Question 6

How many three-digit numbers are divisible by 87?

Solution 6 Question 7

For what value of n, the nth term of A.P 63, 65, 67, …….. and nth term of A.P. 3, 10, 17,…….. are equal to each other?

Solution 7 Question 8

Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

Solution 8 Question 9

If numbers n - 2, 4n - 1 and 5n + 2 are in A.P. find the value of n and its next two terms.

Solution 9 Question 10

Determine the value of k for which k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are in A.P

Solution 10 Question 11

If a, b and c are in A.P show that:

(i) 4a, 4b and 4c are in A.P

(ii) a + 4, b + 4 and c + 4 are in A.P.

Solution 11 Question 12

An A.P consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.

Solution 12 Question 13

4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd time by I. Find the first term and the common difference.

Solution 13 Question 14

The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P

Solution 14 Question 15

In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n - r)

Solution 15 Question 16

Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13th term?

Solution 16 ## Chapter 10 - Arithmetic Progression Exercise Ex. 10(C)

Question 1

Find the sum of the first 22 terms of the A.P.: 8, 3, -2, ………..

Solution 1 Question 2

How many terms of the A.P. :

24, 21, 18, ……… must be taken so that their sum is 78?

Solution 2 Question 3

Find the sum of 28 terms of an A.P. whose nth term is 8n - 5.

Solution 3 Question 4(i)

Find the sum of all odd natural numbers less than 50

Solution 4(i) Question 4(ii)

Find the sum of first 12 natural numbers each of which is a multiple of 7.

Solution 4(ii) Question 5

Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.

Solution 5 Question 6

The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms

Solution 6 Question 7

The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.

Solution 7 Question 8

Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Solution 8 Question 9

The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Solution 9 Question 10

In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.

Solution 10 Question 11

If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.

Solution 11 Question 12

Find the sum of all multiples of 7 lying between 300 and 700.

Solution 12 Question 13

The sum of n natural numbers is 5n2 + 4n. Find its 8th term

Solution 13 Question 14

The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P and the sum of first 50 terms

Solution 14 ## Chapter 10 - Arithmetic Progression Exercise Ex. 10(D)

Question 1

Find three numbers in A.P. whose sum is 24 and whose product is 440.

Solution 1 Question 2

The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms

Solution 2 Question 3

The angles of a quadrilateral are in A.P. with common difference 20o. Find its angles.

Solution 3 Question 4

Divide 96 into four parts which are in A.P and the ratio between product of their means to product of their extremes is 15 :7.

Solution 4 Question 5

Find five numbers in A.P. whose sum is 12(1/2) and the ratio of the first to the last terms is 2: 3.

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.

Question 6

Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.

Solution 6 Question 7

The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.

Solution 7 Question 8

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Solution 8 Question 9

Insert one arithmetic mean between 3 and 13

Solution 9 Question 10

The angles of a polygon are in A.P with common difference 5o. If the smallest angle is 120o, find the number of sides of the polygon

 Let the number sides be n Solution 10 Question 11 Solution 11 ## Chapter 10 - Arithmetic Progression Exercise Ex. 10(E)

Question 1

Two cars start together in the same direction from the same place. The first cargoes at uniform speed of 10 km h-1. The second car goes at a speed of 8 km h-1 in the first hour and thereafter increasing the speed by 0.5 km h-1 each succeeding hour. After how many hours will the two cars meet?

 Let the two cars meet after n hours, then Solution 1 Question 2

A sum of Rs. 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs. 20 less than its preceding prize; find the value of each of the prizes.

Solution 2 Question 3

An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installments is Rs. 100 less than the previous one, find :

(i) amount of installments paid in the 9th month

(ii) total amount paid in the installment scheme

Solution 3 Question 4

A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year find:

(i) the production in the first year.

(ii) the production in the 10th year.

(iii) the total production in 7 years.

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 Question 5

Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying instalments every month. If the instalments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th instalments of loan? What amount of loan she still has to pay after the 30th instalments?

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 Exercise Ex. 10(F)

Question 1

The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.

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

Question 2

If the third and the 9th term of an A.P. be 4 and -8 respectively, find which term is zero?

Solution 2 Question 3

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.

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

Question 4

Find the arithmetic mean of:

(i) -5 and 41

(ii) 3x - 2y and 3x + 2y

(iii) (m + n)2 and (m - n)2

Solution 4 Question 5

Find the sum of first 10 terms of the A.P.

4 + 6 + 8 + ……

Solution 5

Here,

First term, a = 4

Common difference, d = 6 - 4 = 2

n = 10 Question 6

Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.

Solution 6

Here,

First term, a = 3

Last term, l = 57

n = 20 Question 7

How many terms of the series 18 + 15 + 12 + …….. when added together will give 45?

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.

Question 8

The nth term of a sequence is 8 - 5n. Show that the sequence is an A.P.

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.

Question 9

Find the general term (nth term) and 23rd term of the sequence 3, 1, -1, -3, ….. .

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

Question 10

Which term of the sequence 3, 8, 13, ........ is 78?

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.

Question 11

Is -150 a term of 11, 8, 5, 2, ....... ?

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.

Question 12

How many two digit numbers are divisible by 3?

Solution 12 Question 13

How many multiples of 4 lie between 10 and 250?

Solution 13 Question 14

The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the sixth term and the tenth is 44. Find the first three terms of the A.P.

Solution 14 Question 15

The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.

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

Question 16

The 25th term of an A.P. exceeds its 9th term by 16. Find its common difference.

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

Question 17

For an A.P., show that:

(m + n)th term + (m - n)th term = 2 × mthterm

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.

Question 18

If the nth term of the A.P. 58, 60, 62,.... is equal to the nth term of the A.P. -2, 5, 12, …., find the value of n.

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

Question 19

Which term of the A.P. 105, 101, 97 … is the first negative term?

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.

Question 20

How many three digit numbers are divisible by 7?

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.

Question 21

Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.

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.

Question 22

Can 2n2 - 7 be the nth term of an A.P? Explain.

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.

Question 23

Find the sum of the A.P., 14, 21, 28, …, 168.

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.

Question 24

The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31st term of this A.P.

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.

Question 25

Find the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58.

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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