Question

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

Answer 1

Given AP series with first term ,a= 104 and difference d= 13

AP: 104 , 117 , 130 ........... 988

a = 104 , an = 988

988 = 104 + (n-1) 13

988 - 104 = (n-1)13

$\frac{884}{13}=n-1$

$68+1=n$

$\therefore n=69$

$s_n=\frac{n}{2}(a+a_n)$

$S_n=\frac{n}{2}(104+988)$

$S_n=\frac{69}{2}\times \left(1092\right)$

$\therefore S_n=37674$