Question

Find the sum of all three digit natural number which are divisible by 13

Answer 1

First 3 digit natural number divisible by 13= a= 104
c.d= 13
Last 3 digit number divisible by 13= 988
=> Tn = 988
=> a + (n-1) d = 988
=> 104 + (n-1) 13 = 988
=> (n-1) = (988-104)÷ 13
=> n-1 = 68
=> n = 69
Now, Sum of 69 terms
= n/2 (first term + last term)
=69/2 × 1092
= 37674