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

1) 1200
2) 1350
3) 1850
4) 2550
5) 2650
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We get the following numbers
102, 114,126...

a common difference of 12

a1 =102
tn =200
200 = 102 + (n-1) 12
98 =12n -12
n = 110/12 = 9.6
n = 9

sum = 9/2 ( 2*102 + (8)12) = 150 * 9 = 1350

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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