The sum of $n$terms of an$A.P.$ is $3n^2+5$ The number of  term which equals 159, is

Let $S_n$ denote the sum of $n$terms. Then,

$S_n=3n^2+5$

Now,

      $\begin{array}{l}{a}_n=S_n-S_{n-1}\\ a_n=\left(3n^2+5\right)-\left(3(n-1{)}^2+5\right)=6n-3\\ a_n=159\Rightarrow 6n-3=159\Rightarrow 6n=162\Rightarrow n=27.\end{array}$