Question

2. Find the sum of the integers between 100 and 200
that are
not divisible by 9.






​

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

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Answer 2

Let the number of terms between 100 and 200 which is divisible by 9 = n an = a + (n – 1)d 198 = 108 + (n – 1)9 90 = (n – 1)9 n – 1 = 10 n = 11 Sum of an AP = Sn = (n/2) [ a + an] Sn = (11/2) × [108 + 198] = (11/2) × 306 = 11(153) = 1683