Class 10 RD Sharma Solutions – Chapter 9 Arithmetic Progressions – Exercise 9.2

Problem 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, 4,5 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 successive difference of list is same i.e 5

∴ The given sequence is in A.P and have common difference of 5

Problem 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 = 3.1.1 – 5= 3 – 5 = -2

a₂ = 3.2.2 – 5 = 12 – 5 = 7

a₃ = 3.3.3 – 5 = 27 – 5 = 22

a₄ = 3.4.4 – 5 = 48 – 5 = 43

We can see that,

a₂ – a₁ = 7 – (-2) = 9

a₃ – a₂ = 22 – 7 = 15

a₄ – a₅ = 43 – 22 = 21

Since, the successive difference of list is not the same 

∴ The given sequence is not in A.P 

Problem 3: The general term of a sequence is given by a_n = -4n + 15. Is the sequence an A.P.? If so, find its 15th 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 successive difference of list is same i.e -4

∴ The given sequence is in A.P and have common difference of -4

Hence, the 15th term will be

a₁₅ = -4(15) + 15 = -60 + 15 = -45

And, a₁₅ = -45

Problem 4: Write the sequence with nth term :

(i) a_n = 3 + 4n

Solution:

Given:

a_n = 3 + 4n

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

a₁ = 3 + 4.1 = 7 

a₂ = 3 + 4.2 = 11

a₃ = 3 + 4.3 = 15

a₄ = 3 + 4.4 = 19

∴ The sequence is 7, 11, 15, 19

We can see that,

a₂ – a₁ = 11 – (7) = 4

a₃ – a₂ = 15 – (11) = 4

a₄ – a₃ = 19 – (15) = 4

Since, the successive difference of list is same i.e 4

∴ The given sequence is in A.P

(ii) a_n = 5 + 2n

Solution:

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

a₁ = 5 + 2.1 = 7

a₂ = 5 + 2.2 = 9

a₃ = 5 + 2.3 = 11

a₄ = 5 + 2.4 = 13

∴ The sequence is 7, 9, 11, 13

We can see that,

a₂ – a₁ = 9 – (7) = 2

a₃ – a₂ = 11 – (9) = 2

a₄ – a₃ = 13 – (11) = 2

Since, the successive difference of list is same i.e 2

∴ The given sequence is in A.P

(iii) a_n = 6 – n

Solution:

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

a₁ = 6 – 1 = 5

a₂ = 6 – 2 = 4

a₃ = 6 – 3 = 3

a₄ = 6 – 4 = 2

∴ The sequence is 5, 4, 3, 2

We can see that,

a₂ – a₁ = 4 – (5) = -1

a₃ – a₂ = 3 – (4) = -1

a₄ – a₃ = 2 – (3) = -1

Since, the successive difference of list is same i.e -1

∴ The given sequence is in A.P

(iv) a_n = 9 – 5n

Solution:

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

a₁ = 9 – 5.1 = 4

a₂ = 9 – 5.2 = -1

a₃ = 9 – 5.3 = -6

a₄ = 9 – 5.4 = -11

∴ The sequence is 4, -1, -6, -11

We can see that,

a₂ – a₁ = -1 – (4) = -5

a₃ – a₂ = -6 – (-1) = -5

a₄ – a₃ = -11 – (-6) = -5

Since, the successive difference of list is same i.e -5

∴ The given sequence is in A.P

Problem 5: The nth term of an A.P. is 6n + 2. Find the common difference.

Solution:

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

a₁ = 6.1 + 2 = 8

a₂ = 6.2 + 2 = 14

a₃ = 6.3 + 2 = 20

a₄ = 6.4 + 2 = 26

We can see that,

a₂ – a₁ = 14 – (8) = 6

a₃ – a₂ = 20 – (14) = 6

a₄ – a₃ = 26 – (20) = 6

Hence, the common difference is 6

Problem 6: Justify whether it is true to say that the sequence, having following nth term is an A.P.

(i) a_n = 2n – 1

Solution:

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

a₁ = 2.1 – 1 = 1

a₂ = 2.2 – 1 = 3

a₃ = 2.3 – 1 = 5

a₄ = 2.4  – 1 = 7

We can see that,

a₂ – a₁ = 3 – (1) = 2

a₃ – a₂ = 5 – (3) = 2

a₄ – a₃ = 7 – (5) = 2

Since, the successive difference of list is same i.e 2

Hence, the given sequence is in A.P

(ii) a_n = 3n² + 5

Solution:

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

a₁ = 3.1.1 + 5 = 8

a₂ = 3.2.2 + 5 = 17

a₃ = 3.3.3 + 5 = 32

a₄ = 3.4.4 + 5 = 53

We can see that,

a₂ – a₁ = 17 – (8) = 9

a₃ – a₂ = 32 – (17) = 15

a₄ – a₃ = 53 – (32) = 21

Since, the successive difference of list is not the same 

Hence, the given sequence is not in A.P

(iii) a_n = 1 + n + n²

Solution:

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

a₁ = 1 + 1 + 1.1 = 3

a₂ = 1 + 2 + 2.2 = 7

a₃ = 1 + 3 + 3.3 = 13

We can see that,

a₂ – a₁ = 7 – (3) = 4

a₃ – a₂ = 13 – (7) = 6

Since, the successive difference of list is not the same

Hence, the given sequence is not in A.P