Question

Find the sum of all 3 digit natural
numbers, which are divisible by 9.

Answer 1

First 3 digit number which is divisible by 9 is 108 and last one is 999

so , a = 108 and l = 999 , d = 9

According to Arithmetic Progression ,

l = a + (n-1)d

999 = 108 + (n-1)9

891 /9 = n -1

99 =n-1

n = 100

Sum = ( n/2) [ a + l ]

sum = (100/2)( 108 + 999) = 50 × 1107 = 55350