Question

find the sum of first 8 multiples of 3

Answer 1

8th multiple = 3*8=24
a=3 l=24 d=3 n=8
now...
s8= 8/2(2*3+(8-1)3)
=4(6+21)
=4*27

= 108




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Answer 2

Hi friend. Harish here.

Here is your answer:

To find,

The sum of first 8 multiples of 3 .

Solution,

The first 8 multiples of 3 are:

3, 6,9,12,15,18,21,24.

They are in AP as they have same difference:

 (d) = 6 - 3 = 3.

First term of AP (a) = 3.

Total terms n = 8.

Then,

We know that,

$S_n= \frac{n}{2}(2a + (n-1)d)$

→$S_8 = \frac{8}{2} (2\times3 + (8-1) 3)$

→ $S_8 = 4 (6+ 7\times 3)$

→  $4 \times 27$

→ $108$

Therefore their sum is 108.
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Hope my answer is helpful to you.