The sum of the first 12 multiples of 6 is:

Correct option is B

Given:

First term (a) = 6 × 1 = 6

Common difference (d) = 6

Number of terms (n) = 12

Formula Used:

Sum of n terms of an AP

​$$S_n = \frac n2 \times$$​[2a + (n-1)d]

Solution:

Here, a = 6, d = 6, and n = 12.Substitute these values into the formula:

​$$S_{12} = \frac{12}2 \times$$​ [2(6) + (12-1)6]

​$$S_{12} = 6 \times$$​[12 + 66]

​$$S_{12}$$​ = 6 $$\times$$​ [78]

​$$S_{12 }$$​= 468

Alternative Method:

The first 12 multiples of 6 are: 6, 12, 18, ..., 72.

This is an arithmetic progression with the first term 6 and the last term 72.

The sum can also be calculated using the formula:

​$$S_n = \frac n2 \times$$​ (first term + last term)

​$$S_{12} =\frac{ 12}2 \times$$​ (6 + 72)

​$$S_{12} = 6\times$$​ (78)

​$$S_{12}$$​ = 468

The sum of the first 12 multiples of 6 is 468.