In this Question you have to find the sum of all three digit natural number which are divisible by the given number.

This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE Board.

This is the Question Number 12 Of Exercise 11 C of RS Aggarwal Solution

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# Find the sum of all three-digit natural numbers which are divisible by 13.

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The sum of 3-digit number between 100 and 999 that are divisible by 13 can be found out by arithmetic sum i.e.

First 3-digit number that is divided by 13 is 104

Greatest 3-digit number that is divided by 13 is 988

Formula for the sum of Arithmetic progression is n/2(a+l) with “a” being the value of the first number of the series and “l” being the last.

Therefore, a = 104 and l = 988

Value of n depends on the larger number which is divisible by 13 that is 988 by 13 is 76, whereas the number 104 divided by 13 is 8, so the number of terms is 76 – 13 = 69

The sum is n/2(a+l)=69/2(104+988)=37674