Question

find the sum of first 8 multiples if 3

Answer 1

Answer:

Step-by-step explanation:

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

Sum (S) = 3+6+9+12+15+18+21+24

Above series is in Arthematic progression with a=3, d= 3 , l= 24 , n = 8

We know that the sum of arthematic progression. S =(n/2)(a+l)

Where, a= first number in series

l= last number in series

S = (8/2)(3+24)

S = 4(27)

S = 108

Hence sum of the first eight multiples of 3 is 108

(Or)

sum(S)= 3+6+9+12+15+18+21+24

S = 3(1+2+3+4+5+6)

We know that sum of first n natural numbers is

(n/2)(n+1)

Here n=8

Therefore

S=3(8/2)(8+1)

S= 3(4)(9)

S = 12(9)

S = 108

Therefore sum of the first eight multiples of 3 is 108

Answer 2

a=3

n=8

d=3

So,

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

S8=8/2(2*3+(7)(3)

S8=4(27)

S8=108