Question

The sum of those numbers between 100 and 200 which is divisible by 9 will be

Answer 1

Answer:

(i) Number between 100−200 divisible by 9 are 108,117,126,...198

Here, a=108.d=117−108=9 and a

n

=198

=a+(n−1)d=198

→ 108+(n−1)9=198

→ 9[12+n−1]=198⇒ n=22−11⇒ n=11

Now, S

n

=

2

n

[2a+(n−1)d]

⇒ S

11

=

2

11

[2(108)+(11−1)(9)]

=

2

11

[216+90]

=

2

11

×306

=11×153

⇒ S

11

=1683

Step-by-step explanation:

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