RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.2

The second exercise of the chapter is based on understanding the definition and properties of an A.P. The fundamentals of A.P. will be strengthened when a student refers to RD Sharma Solutions Class 10 as it contains detailed solutions to problems aiming to clear doubts. Students can also access the RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.2 PDF provided below for further clarity in solving problems of this exercise.

RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.2

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Access RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progressions Exercise 9.2

1. Show that the sequence defined by a_n = 5n – 7 is an A.P., find its common difference.

Solution:

Given, a_n = 5n – 7

Now putting n = 1, 2, 3, 4, we get,

a₁ = 5(1) – 7 = 5 – 7 = -2

a₂ = 5(2) – 7 = 10 – 7 = 3

a₃ = 5(3) – 7 = 15 – 7 = 8

a₄ = 5(4) – 7 = 20 – 7 = 13

We can see that,

a₂ – a₁ = 3 – (-2) = 5

a₃ – a₂ = 8 – (3) = 5

a₄ – a₃ = 13 – (8) = 5

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = 5n – 7 is an A.P with a common difference of 5.

2. Show that the sequence defined by a_n = 3n² – 5 is not an A.P.

Solution:

Given, a_n = 3n² – 5

Now putting n = 1, 2, 3, 4, we get,

a₁ = 3(1)² – 5= 3 – 5 = -2

a₂ = 3(2)² – 5 = 12 – 5 = 7

a₃ = 3(3)² – 5 = 27 – 5 = 22

a₄ = 3(4)² – 5 = 48 – 5 = 43

We can see that,

a₂ – a₁ = 7 – (-2) = 9

a₃ – a₂ = 22 – 7 = 15

a₄ – a₃ = 43 – 22 = 21

Since the difference between the terms is not common and varying, we can conclude that the given sequence defined by a_n = 3n² – 5 is not an A.P.

3. The general term of a sequence is given by a_n = -4n + 15. Is the sequence an A.P.? If so, find its 15^th term and the common difference.

Solution:

Given, a_n = -4n + 15

Now putting n = 1, 2, 3, 4, we get,

a₁ = -4(1) + 15 = -4 + 15 = 11

a₂ = -4(2) + 15 = -8 + 15 = 7

a₃ = -4(3) + 15 = -12 + 15 = 3

a₄ = -4(4) + 15 = -16 + 15 = -1

We can see that,

a₂ – a₁ = 7 – (11) = -4

a₃ – a₂ = 3 – 7 = -4

a₄ – a₃ = -1 – 3 = -4

Since the difference between the terms is common, we can conclude that the given sequence defined by a_n = -4n + 15 is an A.P with a common difference of -4.

Hence, the 15^th term will be

a₁₅ = -4(15) + 15 = -60 + 15 = -45