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# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression

## Selina Textbook Solutions Chapter 10 - Arithmetic Progression

Selina Textbook Solutions are a perfect way to ace your examination with high marks. These Textbook Solutions are extremely helpful for solving difficult questions in the ICSE Class 10 Mathematics exam. Our Selina Textbook Solutions are written by our subject experts. Find all the answers to the Selina textbook questions of Chapter 10 - Arithmetic Progression.

All Selina textbook questions of Chapter 10 - Arithmetic Progression solutions are created in accordance with the latest ICSE syllabus. These free Textbook Solutions for ICSE Class 10 Selina Concise Mathematics will give you a deeper insight on the fundamentals in this chapter and will help you to score more marks in the final examination. ICSE Class 10 students can refer to these solutions while doing their homework and while studying and revising for the Mathematics exam.

Exercise/Page

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 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

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 10(A)

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 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

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 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 ## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 10(C)

Solution 1 Solution 2 Solution 3 Solution 4(ii) Solution 4(i) Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 ## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 10(D)

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 ## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 10 - Arithmetic Progression Page/Excercise 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 Solution 6 ## Browse Study Material

### Browse questions & answers

TopperLearning provides step-by-step solutions for each question in each chapter. Access Chapter 10 - Arithmetic Progression  for ICSE Class 10 Mathematics free of cost. The solutions are provided by our subject matter experts. Refer to our solutions for the Selina Concise Mathematics textbook to revise the whole chapter and clear your fundamentals before the examination. By referring to the solutions for this chapter and the others, we hope that you are able to write your exams well.

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