Question

Given in an A.P. $a_n=4,d=2,S_n=-14$, find $n$ and $a$.

Answer 1

$a_n=a+(n-1)d=4$

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

$=a+2n-2=4$

$=a+2n=6$

$=a=6-2n$

$S_n=n/2[2a+(n-1\left)d\right]=-14$

$=n/2[2a+(n-1\left)2\right]=-14$

$=an+n(n-1)=-14$

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

$=6n-2n^2+n^2-n=-14$

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

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

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

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

$n=7$ Or $n=-2.$

'n' cannot be negative.

So n = 7.

$\therefore a=6-2\times 7.$

$a=6-14=-8$

$\therefore a=-8$ & $n=7$