find the sum of all two digit natural number which when divided by 3 gives 1 as a remainder
Answers
Answered by
4
According to the problem ,the series is arithmetic with nth term 3n+1
The maximum n for which nth term will be two digit is 10.
The sum is = 3×(n ×(n+1)/2)+n
n=10
So sum = 3×10×11/2 +10
=165+10=175
Similar questions