CAT 2022 Slot 1 QA Question 8: 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 pos

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 pos

Accepted Answer

The $n^{	ext{th}}$ term of a sequence is obtained by subtracting consecutive sums.
$S_n = n + 2n^2$
$S_{n-1} = (n-1) + 2(n-1)^2$
$= n-1 + 2n^2 - 4n + 2$
$= 2n^2 - 3n + 1$
$T_n = S_n - S_{n-1}$
$= (n+2n^2) - (2n^2 - 3n + 1)$
$= 4n - 1$
For this to be divisible by $9$,
$4n - 1 \equiv 0 \pmod{9}$
$\Ri

For any natural number nnn, suppose the sum of the first nnn terms of an arithmetic progression is n+2n2n + 2n^2n+2n2. If the nthn^{\text{th}}nth term of the progression is divisible by 999, then the smallest possible value of nnn is

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