If the sum of $n$ terms of an A.P. is $3 n^{2}+5 n$ then which of its terms is $164 ?$
(a) 26 th
(b) 27 th
(c) 28 th
(d) none of these.
Here, the sum of first n terms is given by the expression,
$S_{n}=3 n^{2}+5 n$
We need to find which term of the A.P. is 164.
Let us take 164 as the nth term
So we know that the nthterm of an A.P. is given by,
$a_{n}=S_{n}-S_{n=1}$
So,
$164=S_{n}-S_{n-1}$
$164=3 n^{2}+5 n-\left[3(n-1)^{2}+5(n-1)\right]$
Using the property,
$(a-b)^{2}=a^{2}+b^{2}-2 a b$
We get,
$164=3 n^{2}+5 n-\left[3\left(n^{2}+1-2 n\right)+5(n-1)\right]$
$164=3 n^{2}+5 n-\left[3 n^{2}+3-6 n+5 n-5\right]$
$164=3 n^{2}+5 n-\left(3 n^{2}-n-2\right)$
$164=3 n^{2}+5 n-3 n^{2}+n+2$
$164=6 n+2$
Further solving for n, we get
$6 n=164-2$
$n=\frac{162}{6}$
$n=27$
Therefore, 164 is the $27^{\text {th }}$ term of the given A.P.
Hence the correct option is (b).