The question asks for the sum of the first 12 multiples of 4. These multiples form a sequence:
This sequence is an arithmetic progression (AP) where:
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 $
The sum of the first 12 multiples of 4 is 312.
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