Discussion

Question:

For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+ 2n²). If the n^th term of the progression is divisible by 9, then the smallest possible value of n is

Explanation:

S_n = n + 2n²

T_n = S_n – S_n-1
⇒ T_n = n + 2n² – [(n-1) + 2(n-1)²]
⇒ T_n = n + 2n² – [n-1 + 2n² – 4n + 2]
⇒ T_n = n + 2n² – n + 1 - 2n² + 4n – 2
⇒ T_n = 4n – 1

T_n is divisible by 9 when n = 7.

Hence, option (d).

» Your doubt will be displayed only after approval.

Doubts

Feedback