Question

find the sum of first 8 multiples of 3

Answer 1

a = 3

d = 3

n = 8

Sn = n/2 [ 2a + (n-1) d]

Required sum = 8 /2 [ 2× 3 + (8 - 1) (3)]

= 4 ( 6 + 21)

= 4 × 27

= 108

Other method :-

First 8 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24

Required sum = 3 + 6 + 9 + 12 + 15 + 18 + 21 +24

= 30 + 15 + 18 + 45

= 45 + 18 +45

= 90 +18

= 108

Answer 2

first 8 multiple of three are........
3,6,9,12,15,18,21,24......

now sum of these numbers is.....

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

hope this helps.....
if correct......
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