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Question

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

A
1659
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B
1693
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C
1653
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D
1683
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Solution

The correct option is B 1683
the series is given by,

108,117,126....198 and so on

So a=108

l=198

n=?

nth term=a+(n1)d

198=108+(n1)9

198=108+9n9

19899=9n

n=11

Now, sum of n terms

Sum of first 11 terms

=112(108+198)

=112(306)

=11×153

=1683

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