Q : 3       In an AP:              (vi) given  $\small a=2,d=8,S_n=90,$ find $\small n$ and $\small a_n$. .

G Gautam harsolia

It is given that
$\small a=2,d=8,S_n=90,$
Now, we know that
$S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}$
$\Rightarrow 90 = \frac{n}{2}\left \{ 2\times(2) +(n-1)8\right \}$
$\Rightarrow 180 = n\left \{ 4+8n-8\right \}$
$\Rightarrow 8n^2-4n-180=0$
$\Rightarrow 4(2n^2-n-45)=0$
$\Rightarrow 2n^2-n-45=0$
$\Rightarrow 2n^2-10n+9n-45=0$
$\Rightarrow (n-5)(2n+9)=0$
$\Rightarrow n = 5 \ \ and \ \ n = - \frac{9}{2}$
n can not be negative so only value of n is 5
Now,
$a_{5} = a+ 4d = 2+4\times 8 = 2+32 = 34$
Therefore, value of n and nth term is 5 and 34 respectively

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