In an AP given a_{n}=4, d=2, S_{n}=-14,find n | vedclass

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(A) Given that,$a_{n}=4, d=2, S_{n}=-14$.Using the formula for the $n^{th}$ term of an $AP$:$a_{n} = a + (n - 1)d$$4 = a + (n - 1)2$$4 = a + 2n - 2$$a

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$n=7, a=-8$

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(A) Given that,$a_{n}=4, d=2, S_{n}=-14$.Using the formula for the $n^{th}$ term of an $AP$:$a_{n} = a + (n - 1)d$$4 = a + (n - 1)2$$4 = a + 2n - 2$$a = 6 - 2n$  --- $(i)$Using the formula for the sum of $n$ terms of an $AP$:$S_{n} = \frac{n}{2}[a + a_{n}]$$-14 = \frac{n}{2}[a + 4]$$-28 = n(a + 4)$Substitute $a = 6 - 2n$ from $(i)$ into the equation:$-28 = n(6 - 2n + 4)$$-28 = n(10 - 2n)$$-28 = 10n - 2n^{2}$$2n^{2} - 10n - 28 = 0$$n^{2} - 5n - 14 = 0$$(n - 7)(n + 2) = 0$Since $n$ must be a positive integer,$n = 7$.Substitute $n = 7$ into equation $(i)$:$a = 6 - 2(7)$$a = 6 - 14$$a = -8$Thus,$n = 7$ and $a = -8$.

In an $AP$ given $a_{n}=4, d=2, S_{n}=-14$,find $n$ and $a$.

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