Find the sum of $3 + 1 - 1 - 3 - 5 - . . .$ to $15$ terms.

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Solution

$S = 3 + 1 - 1 - 3 - 5 - . . .$ to $15$ terms.
$a = 3, d = 1 - 3 = - 2, n = 15$
We know that $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{15} = \frac{15}{2} [2 \times 3 + (15 - 1) \times - 2]$
$S_{15} = \frac{15}{2} [6 - 28]$
$S_{15} = \frac{15}{2} \times - 22 = 15 \times - 11 = - 165$

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