Question

Find the sum of first 8 multiples of 3?

Answer 1

Hey there!

First 8 multiples of 3 = 3 , 6 , ... , 24 .

Sum of first 8 multiples of 3 = 3 + 6 + 9 + .........+ 24 = 3 ( 1 + 2 + ... + 8 )

$\boxed{ Quick \: Formulae : [ \: Sum \: of \: first \: n \: natural \: numbers \: = n(n+1)/2 ] }$

Now,
Sum of first 8 multiples of 3
= 3 [ 1 + 2 + ... + 8 ]
= 3[ 8(9)/2 ]
= 3 [4]9
= 108 .

$\therefore$ Sum of first 8 multiples of 3 = 108

Answer 2

Sum refers to the addition operation.

Multiples of a number is the number which is divisible by the number and by 1.

Hence, the first 8 multiples of three would be,

3*1=3

3*2=6

3=3=9

3*4=12

3*5=15

3*6=18

3*7=21

3*8=24

The multiples are 3,6,9,12,15,18,21,24.

Hence their sum,

3+6+9+12+15+18+21+24= 108

The answer is 108.

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