Aspire Faculty ID #15277 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

Let sum of $n$ terms of an A.P. be $S_n=3n^2+5$. If $T_n$ (the $n$th term) of this series is $159$, then $n$ is …

Solution

$T_n=S_n-S_{n-1}=(3n^2+5)-[3(n-1)^2+5]=6n-3$. Set $6n-3=159 \Rightarrow 6n=162 \Rightarrow n=27$.

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