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 + 4n - 4]$
$ \Rightarrow 90 = n[4n - 2]$
$ \Rightarrow 90 = 2n[2n - 1]$
$ \Rightarrow 45 = n[2n - 1]$
$ \Rightarrow 45 = 2n^2 - n$
$ \Rightarrow 2n^2 - n - 45 = 0$
$ \Rightarrow 2n^2 - 10n + 9n - 45 = 0 $
$ \Rightarrow 2n(n - 5) + 9(n - 5) = 0 $
$ \Rightarrow (n - 5) (2n + 9) = 0$
$ \Rightarrow n - 5 = $0 or 2n + 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 + 4n - 4]$
$ \Rightarrow 90 = n[4n - 2]$
$ \Rightarrow 90 = 2n[2n - 1]$
$ \Rightarrow 45 = n[2n - 1]$
$ \Rightarrow 45 = 2n^2 - n$
$ \Rightarrow 2n^2 - n - 45 = 0$
$ \Rightarrow 2n^2 - 10n + 9n - 45 = 0 $
$ \Rightarrow 2n(n - 5) + 9(n - 5) = 0 $
$ \Rightarrow (n - 5) (2n + 9) = 0$
$ \Rightarrow n - 5 = $0 or 2n + 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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