The sum of first n terms fo an AP is $(5 n - n^{2})$ the nth term of the AP is
(a) (5-2n) (b) (6-2n) (c) (2n-5) (d) (2n-6)

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Solution

Given that sum of the first n terms of an AP is $(5 n - n^{2})$

$S_{n} = (5 n - n^{2})$

First term, $a_{1} = S_{1} = 5 (1) - (1)^{2} = 5 - 1 = 4$

Sum of first two terms $S_{2} = 5 (2) - (2)^{2} = 10 - 4 = 6.$

Second term, $a_{2} = S_{2} - S_{1} = 6 - 4 = 2$

$d = a_{2} - a_{1} = 2 - 4 = - 2$

$a_{n} = a + (n - 1) d$

$= 4 + (n - 1) (- 2)$

$= 4 - 2 n + 2$

$= 6 - 2 n .$

∴ nth term of an A.P = 6 - 2n.
option b is correct

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