Aspire Faculty ID #15280 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

If sum of $n$ terms of a series is $3n^2 + 4n$, then the series is …

Solution

$S_n = 3n^2 + 4n$ Then, $T_n = S_n - S_{n-1} = (3n^2 + 4n) - [3(n-1)^2 + 4(n-1)] = 6n + 1$. Since $T_n$ is linear in $n$, the series is an A.P.
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