Question

find the sum of integers between 100 and 200 that are divisible by 9

Answer 1

first term (a) = 108 difference {d) = 9. last term = 198. an = a+(n-1 ) d , 198 = 108 + ( n - 1) 9 , solve this , n= 11. sn = n/2 ( 2a + (n-1)d ). sn =11/2 ( 2( 108) + (11-1)9). solve this , sn= 1683

Answer 2

Hey dear,

here is your answer,

First term(a)=108
Common difference(d)=9
Last term(l)=198
No. of term=n

l=an+(n-1)d
198=108+(n-1)9
198-108=9(n-1)
90=9(n-1)
n-1=90/9
n-1=10
n=10+1=11

Sn=n/2[2a+(n-1)d]
Sn=11/2[2*108+(11-1)9]
Sn=11/2[216+90]
Sn=(11/2)*306
Sn=11*153
Sn=1683
 
Hope this will help you out !!
Please mark as brainliest !!