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
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow 90 = \frac{n}{2}\left[ {2(2) + (n - 1)8} \right]$
$ \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$ or $n = - \frac{9}{2}$
$n = - \frac{9}{2}$ is in admissible as $n,$ being the number of terms, is a natural number.
$\therefore n = 5$
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
${S_n} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]$
$ \Rightarrow 90 = \frac{n}{2}\left[ {2(2) + (n - 1)8} \right]$
$ \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$ or $n = - \frac{9}{2}$
$n = - \frac{9}{2}$ is in admissible as $n,$ being the number of terms, is a natural number.
$\therefore n = 5$
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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