In an AP,given a=2, d=8, S_{n}=90,find n and | vedclass

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(A) Given that,$a=2, d=8, S_{n}=90$.Using the formula for the sum of $n$ terms of an $AP$:$S_{n} = \frac{n}{2}[2a + (n-1)d]$$90 = \frac{n}{2}[2(2) + (

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$n=5, a_{n}=34$

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(A) Given that,$a=2, d=8, S_{n}=90$.Using the formula for the sum of $n$ terms of an $AP$:$S_{n} = \frac{n}{2}[2a + (n-1)d]$$90 = \frac{n}{2}[2(2) + (n-1)8]$$90 = \frac{n}{2}[4 + 8n - 8]$$90 = \frac{n}{2}[8n - 4]$$90 = n(4n - 2)$$4n^{2} - 2n - 90 = 0$Dividing by $2$:$2n^{2} - n - 45 = 0$$2n^{2} - 10n + 9n - 45 = 0$$2n(n - 5) + 9(n - 5) = 0$$(n - 5)(2n + 9) = 0$Since $n$ must be a positive integer,$n = 5$.Now,find $a_{n}$:$a_{n} = a + (n-1)d$$a_{5} = 2 + (5-1)8$$a_{5} = 2 + 32 = 34$.Thus,$n = 5$ and $a_{n} = 34$.

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

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