Question

A sequence is such that the sum of its any number of terms, beginning from the first, is four times as large as the square of the number of terms. If the $n^{\text{th }}$ term of such a sequence is 996 , then value of $n$ is equal to

• A. 100
• B. 112
• C. 125
• D. 132

Answer 1

Answer: (C)

Given that $S_n=4n^2$
$\therefore T_n=S_n-S_{n-1}=4n^2-4(n-1{)}^2=4\left[n^2-\left(n^2+1-2n\right)\right]=4(2n-1)=8n-4$ Hence $T_1,T_2,T_3,T_4,\dots \dots \dots ...$. are in A.P.
i.e., $4,12,20,28,\dots \dots .$.
Now, $T_n=996=8n-4\Rightarrow n=125$.