Let there be an A.P. with first term ‘$a$’, common difference '$d$'. If $a_n$ denotes its $n^{th}$ term and $S_n$ the sum of first $n$ terms, find.

$n$ and $a$, if $a_n = 4, d = 2$ and $S_n = -14$.

Given:

In an A.P., first term $=a$ and common difference $=d$.

$a_n$ denotes its $n^{th}$ term and $S_n$ the sum of first $n$ terms.

To do:

We have to find $n$ and $a$, if $a_n = 4, d = 2$ and $S_n = -14$.

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 $n=7$.