In an AP, given a $= 2$ , d $= 8$ , $S_{n} = 90$ , find n and $a_{n}$ .

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Solution

$a = 2, d = 8, s_{n} = 90$
$s_{n} = \frac{n}{2} [2 a + (n - 1) d] 90 = \frac{n}{2} [4 + (n - 1) 8]$
$180 = n [8 n - 4]$
$180 = 8 n^{2} - 4 n$
$= 8 n^{2} - 4 n - 180 = 0$
$= 2 n^{2} - n - 45 = 0$
$= 2 n^{2} - 10 n + 9 n - 45 = 0$
$= 2 n (n - 5) - 9 (n - 5)$
$= (2 n - 9) (n - 5)$
$n = 5$
$a_{n} = a + (n - 1) d a_{n} = 2 + (5 - 1) 8 a_{n} = 2 + 32 a_{n} = 34$

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