What is the sum of the first 12 multiples of 4?

Correct option is B

Given:

We are to find the sum of the first 12 multiples of 4.​

Formula Used:
Sum of first n terms of A.P.:

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

Solution:

We are to find the sum of the first 12 multiples of 4.
These multiples are: 4, 8, 12, $$\dots, 48$$​​

This is an arithmetic progression (A.P.) where:

First term a = 4

Common difference d = 4

Number of terms n = 12

​$$S_{12} = \dfrac{12}{2} \left[2 \times 4 + (12 - 1) 4\right] = 6 \left[8 + 44\right] = 6 \times 52 = 312$$​