Question

Given $a_n=4,d=2,S_n=-14,$ then find the $n$ and $a_n$.

Answer 1

Given. $a_n=4$

$d=2$

$s_n=-14.$

Now, $a_n=4$

$\Rightarrow a_1+(n-1)d=4$ where $a_1$ The first term

$\Rightarrow a_1+(n-1)2=4$

$a_1+2(n-1)=4$ ......(1)

and $s_n=-14$

$\Rightarrow \frac{n}{2}\{2a_1+(n-1\left)d\right\}=-14.$

$\Rightarrow \frac{n}{2}\{2a_1+2(n-1\left)\right\}=-14.$

$\Rightarrow n{a}_1+n(n-1)=-14$

$\Rightarrow n(a_1+(n-1\left)\right)=-14.$

$\Rightarrow (4-2(n-1)+(n-1\left)\right)=-14$

$\Rightarrow (4-2(n-1\left)\right)=-14$ [from (1)]

$\Rightarrow n(4-(n-1\left)\right)=-14$

$\Rightarrow 4n-n(n-1)=-14$

$\Rightarrow 4n-n^2+n=-14$

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

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

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

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

$\Rightarrow n=-2$ or $n=7.$

here n=-2 is rejected since n is not negative.

$\therefore n=7.$

$from\left(1\right),a_1+2(n-1)=4$

$\Rightarrow a_1+2(7-1)=4$

$\Rightarrow a_1+2.6=4$

$\Rightarrow a_1=4-12.$

$\Rightarrow a_1=-8.$

Hence $a_n=a_1+(n-1)d$

$=-8+(7-1)2$

$=-8+6.2$

$=-8+12.$

$=4$