The $n^{t h}$ terms of an A.P. is given by $(- 4 n + 15)$ . Find the sum of first $20$ terms of this A.P.

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Solution

Given: $n^{t h}$ term of AP is $a_{n} = (- 4 n + 15)$
Put $n = 1, 2, 3, . . . . .$ , we get,
$- 4 (1) + 15, - 4 (2) + 15, - 4 (3) + 15, . . . . . . . .$
$= - 4 + 15, - 8 + 15, - 12 + 15, . . . . . . .$
$= 11, 7, 3, . . . . . .$
So, first term $a = 11$ and common difference $d = a_{2} - a_{1} = 7 - 11 = - 4$
Thus, the sum of first $20$ terms of given AP = $S_{20} = \frac{20}{2} (2 (11) + (20 - 1) \times (- 4))$
$= 10 (22 + (19) (- 4))$
$= 10 (22 - 76)$
$= 10 \times (- 54)$
$= - 540$

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