What is the sum of the integers from 100 to 200 inclusive which are divisible by 6 but not by 12?

If you observe the pattern of the numbers that are divisible by 6 but not by 12 in the given range, they lie at a distance of 12 from each other consecutively.

So the series of the terms starts at 102 and ends at 198, they form an AP with common difference of 12.

\(t_{n}=a+(n-1)*d=102+(n-1)*12\)
Or, 198=102+(n-1)*12
Or, n=9

Sum, \(S_{9}=\frac{n}{2}*(a+l)\)=\(\frac{9}{2}*(102+198)=\frac{9}{2}*300=9*150=1350\)

Ans. (B)
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PKN

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This means we need to find the sum of the odd multiples of 6 from 100 to 200, inclusive, since even multiples of 6 will be divisible by 12. The first odd multiple of 6 within the range is 6 x 17 = 102, the next one is 6 x 19 = 114, etc. The last one is 6 x 33 = 198. So the odd multiples of 6 within the range are: 102, 114, 126, …, 198.

This is an arithmetic sequence of integers with a common difference of 12, so the number of integers is

(198 - 102)/12 + 1 = 96/12 + 1 = 9

The average of the integers is (198 + 102)/2 = 300/2 = 150. Thus, the sum is

150 x 9 = 1350

Answer: B/2
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