Question

find the sum of first 12 multiples of 6​

Answer 1

Hello !

Well, we have that the first 12 multiples of 6 are.

0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 and 66.

Now, that you have the first 12 multiples of 6 you can find the sum of all terms using the following method below.

Sn = (n/2) × (A1 + An)

S12 = (12/2) × (0 + 66)

S12 = 6 × 66

S12 = 396

Therefore, we have that the sum of the first 12 multiples of 6 is 396.

I hope I have collaborated !

Answer 2

Answer:

The correct answer is $476.$

Step-by-step explanation:

Given:

the sum of the first 12 multiples of 6​.

To find the sum of the first 12 multiples of 6​.

The sum of the first $12$ multiples of $6= 6,12,18, ........., 72.$

Let, $\\ n &=12, d=6$

Sum of n terms,

$S_{n} &=\frac{n}{2}[2 a+(n-1) d]$

Sum of 12 terms,

$S_{12} &=\frac{12}{2}[2(6)+11 d]$

$&=6[12+11 d]$

$=6[12+11(6)]$

$&=6[12+66]$

$&=6[78]$

$&=476 \end{aligned}$

Therefore, the sum of the first $12$ multiples  $6$ is $476.$

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