Question

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

• A. $\frac{n}{2}(13-3n)$
• B. $\frac{n}{2}(30-3n)$
• C. $\frac{n}{2}(13+3n)$
• D. $n(8-3n)$

Answer 1

The correct option is A $\frac{n}{2}(13-3n)$

In the given A.P first term, a = 5
Common difference, d = 2 - 5 = -3
Now sum of n terms $S_n$ is given by:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]=\left(\frac{n}{2}\right)\left[2\right(5)+(n-1\left)\right(-3\left)\right]=\frac{n}{2}[10-3n+3]=\frac{n}{2}[13-3n]$