Question

the sum of all natural number between 100 to 200 which are multiple of 3

Answer 1

Hey there,

Numbers which are divisible by 3 from 100 are 102, 105....

Sum = 102+105 +......+ 198

= 3(34+35...... + 66)

= 3[( 1+2+3+4....+66) - 1+2+....33]

= 3 [ 66(67)/2 - 33(34) /2

= 3(2211- 561)

= 3(1650)

= 4950

The sum of first ' n ' natural numbers is n( n+1) /2

Hope helped!