Question

In an $A.P.S_n=5n^2+11n,T_n=?$

Answer 1

We have,

$S_n=5n^2+11n$

$\text{On}\text{multiplying}\text{and}\text{divide}\text{by}\text{2}\text{and}\text{we}\text{get,}$

$\Rightarrow S_n=\frac{2}{2}(5n^2+11n)$

$\Rightarrow S_n=\frac{n}{2}(10n+22)$

$\Rightarrow S_n=\frac{n}{2}(22+10n)$

$\Rightarrow S_n=\frac{n}{2}(22+10n-10+10)$

$\Rightarrow S_n=\frac{n}{2}(32+10(n-1))$

$\Rightarrow S_n=\frac{n}{2}(32+(n-1)10)$

On comparing that,

$S_n=\frac{n}{2}(2a+(n-1)d)$

Now,

$a=16,d=10$

So,

$T_n=a+(n-1)d$

$T_n=16+(n-1)10$

$T_n=16+10n-10$

$T_n=6+10n$

Hence, this is the answer.