Question

Given $a=2,d=8$, $S_n$ $=90$, find $n$ and $a_n$

Answer 1

Given a=2.

d=8

$S_n=90$

$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=90$

$\Rightarrow \frac{n}{2}(2.2+(n-1\left)8\right)=90$

$\Rightarrow \frac{n}{2}(4+8(n-1\left)\right)=90$

$\Rightarrow \frac{n}{2}(4+8n-8)=90$

$\Rightarrow \frac{n}{2}(8n-4)=90$

$\Rightarrow \frac{n}{2}(8n-4)=90$

$\Rightarrow 4n^2-2n=90$

$\Rightarrow 2n^2-n=90$

$\Rightarrow 2n^2-n=45$

$\Rightarrow 2n^2-n-45=0$

$\Rightarrow 2n^2-10n-9n-45=0$

$\Rightarrow 2n(n-5)9(n-5)=0$

$\Rightarrow (2n+9)(n-5)=0$

$\Rightarrow m=\frac{-9}{2}$ or $n=5$

Hence n=-9/2 is neglect since n is not negative

$\therefore$ n=5

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

$a_5=a+(5-1)d$

=2+4.8

=2+32

=34