Question
In an AP: $a = 2, d = 8, S_n= 90,$ find $n$ and $a_n$.
✓
Answer
Here,$ a = 2$
$ d=8 $
$ S_n=90$
We know that
$ S_n=\frac{n}{2}[2 a+(n-1) d] $
$ \Rightarrow 90=\frac{n}{2}[2(2)+(n-1) 8] $
$ \Rightarrow 90=n[2+(n-1) 4] $
$ \Rightarrow 90=n[2+4 n-4] $
$ \Rightarrow 90=n[4 n-2] $
$ \Rightarrow 90=2 n[2 n-1] $
$ \Rightarrow 45=n[2 n-1] $
$ \Rightarrow 45=2 n^2-n $
$ \Rightarrow 2 n^2-n-45=0 $
$ \Rightarrow 2 n^2-10 n+9 n-45=0 $
$ \Rightarrow 2 n(n-5)+9(n-5)=0 $
$ \Rightarrow(n-5)(2 n+9)=0 $
$ \Rightarrow n-5=0 \text { or } 2 n+9=0 $
$ \Rightarrow n=5 \text { or } n=-\frac{9}{2}$
$n=-\frac{9}{2}$ is in admissible as $n$, being the number of terms, is a natural number.
$\therefore \mathrm{n}=5$
Again, we know that
$ a_n=a+(n-1) d $
$ \Rightarrow a_n=2(5-1) 8 $
$ \Rightarrow a_n=2+(4) 8 $
$ \Rightarrow a_n=2+32 $
$ \Rightarrow a_n=34$
$ d=8 $
$ S_n=90$
We know that
$ S_n=\frac{n}{2}[2 a+(n-1) d] $
$ \Rightarrow 90=\frac{n}{2}[2(2)+(n-1) 8] $
$ \Rightarrow 90=n[2+(n-1) 4] $
$ \Rightarrow 90=n[2+4 n-4] $
$ \Rightarrow 90=n[4 n-2] $
$ \Rightarrow 90=2 n[2 n-1] $
$ \Rightarrow 45=n[2 n-1] $
$ \Rightarrow 45=2 n^2-n $
$ \Rightarrow 2 n^2-n-45=0 $
$ \Rightarrow 2 n^2-10 n+9 n-45=0 $
$ \Rightarrow 2 n(n-5)+9(n-5)=0 $
$ \Rightarrow(n-5)(2 n+9)=0 $
$ \Rightarrow n-5=0 \text { or } 2 n+9=0 $
$ \Rightarrow n=5 \text { or } n=-\frac{9}{2}$
$n=-\frac{9}{2}$ is in admissible as $n$, being the number of terms, is a natural number.
$\therefore \mathrm{n}=5$
Again, we know that
$ a_n=a+(n-1) d $
$ \Rightarrow a_n=2(5-1) 8 $
$ \Rightarrow a_n=2+(4) 8 $
$ \Rightarrow a_n=2+32 $
$ \Rightarrow a_n=34$
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