cat 2022 Question 18

Question

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

Accepted Answer

It is given,

$S_n = 2n^2 + n$

$S_{n-1} = 2(n - 1)^2 + (n - 1)$

$S_{n-1} = 2n^2 - 3n + 1$

$T_n = S_n - S_{n-1} = 2n^2 + n - 2n^2 + 3n - 1 = 4n - 1$

$T_n = 4n - 1$

The terms are 3, 7, 11, 15, 19, 23, 27,……

27 is the first term in the series divisible by 9.

27 is the 7th term.

Therefore, the least possible value of n is 7.

Question 18.

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Question 18.

Question Explanation

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