Question

find the sum of integers bw 100 and 200 divusible by 9 ​

Answer 1

integer divisible by 9 are

108,117,126.......198

a = 108, d = 9 , an = 198

an = a + (n-1)d = 198

108+ (n-1)9 = 198

(n-1)9 = 90

n-1 = 10

n=10+1 =11

total terms are 11 which are divisible by 9

now,

S11 = 11/2[a+an]

S11 = 11/2[108+198]

S11 = 11/2[306]

S11 = 11*153 = 1683

S11.= 1683

i hope it will help u