Question

What is the sum of first 30 multiples of 4?

• A. 1880
• B. 1860
• C. 1800
• D. 1660

Answer 1

The correct option is B

1860

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

where $S_n$ is the sum of n terms of the AP,

'n' is the number of terms,

'a' is the first term,

'd' is the common difference.

Here, a = 4, n = 30, d = 4

$\therefore$$S_{30}=\frac{30}{2}[2\times 4+(30-1\left)4\right]$ =15[8+116]

$\Rightarrow$$S_{30}$ = 1860