Question

find the sum of all natural numbers from 30 to 130 which is divisible by 3

Answer 1

Hiii friend,

AP = 30,33,36,39....129

Here,

First term (A) = 30

Common difference (D) = 3

Last term (L) = 129

A+(N-1) × D = 129

30 + (N-1) × 3 = 129

30 + 3N -3= 129

3N = 129-27

3N = 102

N = 102/3 = 34

Sn = N/2 × [2A+ (N-1) × D]

S34 = 34/2 × [ 2× 30 + (34-1) × 3]

S34 = 17 × (60 + 99) => 17 × 159

=> 2703.

Hence,

The sum of all natural Number between 30 to 130 which is divisible by 3 is 2703.

HOPE IT WILL HELP YOU..... :-)