Question

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

Accepted Answer

C. Given sum of first $n$ terms $S_n = n + 2n^2$. The $n^{	ext{th}}$ term $T_n = S_n - S_{n-1}$: $$T_n = (n + 2n^2) - ((n-1) + 2(n-1)^2)$$ $$T_n = n + 2n^2 - (n - 1 + 2n^2 - 4n + 2)$$ $$T_n = n + 2n^2 - (2n^2 - 3n + 1) = 4n - 1$$ We need $T_n = 4n - 1$ to be divisible by 9. $$4n - 1 \equiv 0 \pmod 9$$ $$4n \equiv 1 \pmod 9$$ Multiplying both sides by 7 (since $4 	imes 7 = 28 \equiv 1 \pmod 9$): $$n \equiv 7 \pmod 9$$ The smallest natural number $n$ satisfying this is $n = 7$.

Smallest n for Divisibility in AP

Explanation

Practise this under exam conditions