We have asked to find the sum of all natural numbers less than 100 which are divisible by 6.
Sir please give me a detailed solution of this question as it is very important for examination
ML Aggarwal Avichal Publication arithmetic progression chapter 9 question no 20 iv
6,12,18,…96 are all natural less than 100 and divisible by 6
first term =6
common difference= d = a2 -a1=12-6 =6
nth term = 96
a+(n-1)d=96
6+(n-1)6=96
(n-1)6=96-6
(n-1)6=90
n-1=90/6
n-1=15
n=15+1
n=16
sum = n/2[a+ nth term]
=16/2[6+96]
=8×102
=816