What is the sum of the integers from 100 to 200 inclusive

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

If you're unfamiliar with the formulas for arithmetic series, you can also answer the question this way...

We want the sum: 102 + 114 + 126 + . . . + 186 + 198
If we add the terms in PAIRS (starting at each end), we get: (102 + 198 ) + (114 + 186 ) + . . .
Simplify to get: 300 + 300 + ....

The question now is, "How many terms are in the sum 102 + 114 + 126 + . . . 186 + 198?

Notice that 102 = (17)(6) = (2 x 8 + 1)(6)
and 114 = (19)(6) = (2 x 9 + 1)(6)
and 126 = (21)(6) = (2 x 10 + 1)(6)
.
.
.
and 126 = (33)(6) = (2 x 16 + 1)(6)

We can see that the number of terms = the number of integers from 8 to 16 inclusive

A nice rule says: the number of integers from x to y inclusive equals y - x + 1

So, the number of terms = 16 - 8 + 1 = 9
So, there are 9 terms, which means we have 4 pairs that add to 300, and 1 value, which we can deduce must be HALF of 300

So, the SUM = (4)(300) + 150 = 1350

Answer: B

Cheers,
Brent

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