Question

For an $A.P$, if $T_1=22,T_n=-11$ and $S_n=66$, then find $n$.

Answer 1

Given that,

$a=22=t_1$

$\begin{array}{c}{t}_n=22+\left(n-1\right)d=-11\\ \Rightarrow \left(n-1\right)d=-\mathrm{33...........}\left(i\right)\\ S_n=66=\frac{n}{2}\left(2\times a+\left(n-1\right)d\right)\\ \Rightarrow 132=n\left(44+\left(n-1\right)d\right)\end{array}$

$\Rightarrow 132=n\left(44-33\right)$

$\Rightarrow 132=n\left(11\right)$

$n=12$

Then,

We get n term is $12$