Question

the sum of all natural numbers between 100 and 200 which are multiplies of 9 is

Answer 1

first multiple of 9 after 100 is 108
a=108. last multiple of 9 btw 100 and 200
is 198.
so An = 198
and d = 9 (multiples of 9)
An = a + (n-1)d
198 = 108 +(n-1) 9
198-108= (n-1)9
90/9=n-1
10=n-1
=> n = 11
So, Sn = n/2(a + An)
S = 11/2(108+198)
S = 11/2(306) = 11*153 = 1683

Answer 2

First number which is multiple of 9 after 100 is 108 and last no is 198
Nth no = 108+(n-1)9=198
=>9n-9=198-108
=>9n=90+9
=>n=99/9=11
Sum=9/2(108+198)
=11/2×306
=11×153
=1683