In an AP, if $ S_{n}=3 n^{2}+5 n $ and $ a_{k}=164 $, find the value of $ k $.

Given:

In an AP, \( S_{n}=3 n^{2}+5 n \) and \( a_{k}=164 \)

To do:

We have to find $k$.

Solution:

Let $a$ be the first term and $d$ be the common difference.

Let us substitute $n=1, 2$ to find the values of $a$ and $d$

$S_1=3(1)^2+5(1)$

$=3+5$

$=8$

$\Rightarrow a_1=a=8$

$S_2=3(2)^2+5(2)$

$=12+10$

$=22$

Second term $a_2=S_2-S_1$

$=22-8$

$=14$

Therefore,

$d=a_2-a_1$

$=14-8$

$=6$

We know that,

$n$th term $a_n=a+(n-1)d$

$a_k=a+(k-1)d$

$164=8+(k-1)6$

$164-8=(k-1)6$

$156=(k-1)6$

$k-1=26$

$k=26+1$

$k=27$

Therefore, $k=27$.