Question

find the sum of all three digit numbers which are multiples of 9??????​

Answer 1

Answer:

From100to 999 the multiples of 9=109,118,127,136,145,154,163,172,181,190,199,.....so on

looks like there's an AP

with a=109 difference =9

n=11×9=99

so sum of all multiples of 9(of 3 digits) is

Sn=n/2[2a+(n-1)d]

=99/2(2×109+98×9)

=99/2(218+882)

=99/2×1100=54450

hope it helps you thanks