find the sum of the integers between 100 and 200 that are (1)divided by 9 (2) not divided by 9
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Answer:
108,117,126.... 198 and so on
So a=108
l=198
n= ?
nth term= a + (n-1)d
198= 108+(n-1)9
198= 108+9n-9
198-99= 9n
99/9= n
n= 11
Now, sum of n terms
Sum of first 11 terms= 11/2(108+198)
=11/2(306)
=11×153
=1683
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