Question

Justify whether it is true to say that the sequence having nth term, an = 3n2 + 5 is an A.P.

Answer 1

Step-by-step explanation:

Yes, here an = 2n - 3

Put n = 1, a1 = 2(1) -3 = -1

Put n = 2, a2 = 2(2) -3 = 1

Put n = 3, a3 = 2(3) -3 = 3

Put n = 4, a4 = 2(4) -3 = 5

List of numbers becomes -1,1,3,...

Here, a2 - a1 = 1 - (-1) = 1 + 1 = 2

a3 - a2 = 3 - 1 = 2

a4 - a3 = 5 - 3 = 2

∵ a2 - a1 = a3 - a2 = a4 - a3 = ...

Hence, 2n - 3 is the nth term of an AP.

(ii) No, here an = 3n2 + 5

Put n = 1, a1 = 3(1)2 + 5 = 8

Put n = 2, a2 = 3(2)2 + 5 3(4) + 5 = 17

Put n = 3, a3 = 3(3)2 + 5 3(9) + 5 = 27 + 5 = 32

So, the list of number becomes 8,17,32,...

Here, a2 - a1 = 17 - 8 = 9

a3 - a2 = 32 - 17 = 15

∴ a2 - a1 ≠ a3 - a2

Since, the successive difference of the list is not same. So, it does not form an AP.

(iii) No, here an = 1 + n + n2

Put n = 1, a1 = 1 + 1 + (1)2 = 3

Put n = 2, a2 = 1 + 2 + (2)2 = 1 + 2 + 4 = 7

Put n = 3, a3 = 1 + 3 + (3)2 = 1 + 3 + 9 = 13

So, the list of numbers becomes 3,17,13,...

Here, a2 - a1 = 7 - 3 = 4

a3 - a2 = 13 - 7 = 6

∴ a2 - a1 ≠ a3 - a2

Since, the successive difference of the list is not same. So, it does not form an AP.