The sum to n terms of the AP 5, 2, -1, -4, -7, ... is

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Solution

The correct option is A
The sum of $n$ terms of an A.P with first term as $a$ and common difference $d$ is

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

Here, $a = 5$ and $d = 2 - 5 = - 3$

$= (\frac{n}{2}) [2 (5) + (n - 1) (- 3)] = \frac{n}{2} [10 - 3 n + 3] = \frac{n}{2} [13 - 3 n]$

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Similar questions

Find the sum to $n$ terms of the AP:
5, 2, -1, -4, -7, ...

\frac{n}{2} (13 - 3 n)

n

5, 2, - 1, - 4, - 7, . . . . .

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