In an AP:
Given $a_n = 4, d = 2, S_n = -14$, find $n$ and $a$.

Given:

In an A.P., $a_n = 4, d = 2, S_n = -14$

To do:

We have to find $n$ and $a$.

Solution:

We know that,

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

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

This implies,

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

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

$4-2n+2=a$

$a=6-2n$......(i)

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

$-14=\frac{n}{2} \times 2[a+(n-1)]$

$-14=n(6-2n+n-1)$

$-14=n(5-n)$

$-14=5n-n^2$

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

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

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

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

$n=7$ or $n=-2$ which is not possible as $n$ cannot be negative

$\therefore a=6-2(7)$

$=6-14$

$=-8$

Therefore, $a=-8$ and $n=7$.