Question

Find the sum of all the positive integers between 100 and 200 that are divisible by 9.

Answer 1

after 100 the first integer is 108 and the last is 198
the the series-108,117.........198
in this series a=108 d=9 an=198 then we first found out the n
an=a+(n-1)d
198=108+(n-1)9
then n=90/9+1=11
then the sum is sn= n/2[2a+(n-1)d]
sn=11/2[108+198]
sn=11×306/2=11×153=1683
I hope this is right

Answer 2

first number divisible by 9 between 100 and 200= 108
second is= 117
@nd third is 126
and last number divisible by 9 between 100 and 200=198
so we got an AP series
108,117,126................198
a= 108 , d= 9
last term = 198
an = a + (n-1)d
198=108+ (n-1)d
(n-1)×9=90
n-1 = 10 or n=11
so sum = n÷ 2 (a+l)
s11= 11÷2(108+198)
s11= 11÷2(306)
s11=11(153)
s11=1683