Question

In an $A.P.$, $a_n=4,d=2,s_n=-14$, find $n$ and $a$.

Answer 1

As we know that, in an A.P.-

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

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

Given:-

$a_n=4$

$S_n=-14$

$d=2$

Now,

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

$4=a+(n-1)2$

$a+2n=4+2$

$\Rightarrow a=6-2n.....\left(1\right)$

Again,

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

$\Rightarrow S_n=\frac{n}{2}[a+a_n]$

$\Rightarrow -14=\frac{n}{2}(a+4)$

$\Rightarrow -14=\frac{n}{2}(6-2n+4)$

$\Rightarrow -14=n(5-n)$

$\Rightarrow -14=5n-n^2$

$\Rightarrow n^2-5n-14=0$

$\Rightarrow n^2-7n+2n-14=0$

$\Rightarrow (n-7)(n+2)=0$

$\Rightarrow n=7,-2$

$\because n\ne -ve$

$\therefore n=7$

Substituting the value of $n$ in equation $\left(1\right)$, we get

$a+6-2n=6-2\times \left(7\right)=6-14=-8$