Find the sum of the integers between 100 and 200 that are . . .

Question : Find the sum of the integers between 100 and 200 that are 

(i) divisible by 9

(ii) Not divisible by 9.

Doubt by Yathartha

Solution : 

i)

Integers between 100 and 200 which are divisible by 9 are 

108, 117, 126, . . .198

a = 108

d= a₂-a₁
d=117-108

d=9

a_n=198

a+(n-1)d=198

108+(n-1)9=198

(n-1)9=198-108
(n-1)9=90

n-1=90/9

n-1=10

n=10+1
n=11

S_n=(n/2)[a+an]

S₁₁=(11/2)[108+198]
S₁₁=(11/2)[306]
S₁₁=11[153]
S₁₁=1683

ii) 

Integers between 100 and 200 are 
101, 102 . . .199

a=101
d=a2-a1

d=102-101
d=1

a_n=199

a+(n-1)d=199

101+(n-1)1=199

n-1=199-101

n-1=98
n=98+1
n=99

Sum of all the integers between 100 and 200 are

S₉₉=(99/2)[a+an]
S₉₉=(99/2)[101+199]

S₉₉=(99/2)[300]
S₉₉=99×150

S₉₉=14850

Sum of all the integers between 100 and 200 which are not divisible by 9 = [Sum of all the integers between 100 & 200] - [Sum of all the integers between 100 & 200 which are divisible by 9]

= 14850 - 1683

= 13167