In this Question you have to find
(i) sum of all natural numbers
(ii) which is divisible by the given digit
This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE Board.
This is the Question Number 9 Of Exercise 11 C of RS Aggarwal Solution
Deepak BoraNewbie
Find the sum of all natural numbers less than 100 which are divisible by 6.
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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