Identifying the Sequence of Multiples

The question asks for the sum of the first 12 multiples of 4. These multiples form a sequence:

• 4 x 1 = 4
• 4 x 2 = 8
• 4 x 3 = 12
• ...
• 4 x 12 = 48

This sequence is an arithmetic progression (AP) where:

• The first term ($a$) is 4.
• The common difference ($d$) is 4.
• The number of terms ($n$) is 12.
• The last term ($l$) is 48.

Calculating the Sum using AP Formula

The sum ($S_n$) of an arithmetic progression can be calculated using the formula:

$ S_n = \frac{n}{2}(a + l) $

Substitute the values:

$ S_{12} = \frac{12}{2}(4 + 48) $

$ S_{12} = 6 \times 52 $

$ S_{12} = 312 $

Alternatively, using the formula $S_n = \frac{n}{2}(2a + (n-1)d)$:

$ S_{12} = \frac{12}{2}(2 \times 4 + (12-1) \times 4) $

$ S_{12} = 6(8 + 11 \times 4) $

$ S_{12} = 6(8 + 44) $

$ S_{12} = 6 \times 52 $

$ S_{12} = 312 $

Final Answer

The sum of the first 12 multiples of 4 is 312.