I dont have the MGMAT, solution so not sure how they did it. But here is what I did -

This is a arithmetic progression. So I first used the forumla for AP to find the 13th and the 28th term. Formula is on the site below -

https://library.thinkquest.org/C0110248/ ... parith.htm

The 13th terms is 78 and the 28th term is 168

Sum of AP is also given on the above site. Using that formula u get Sum of 16 numbers (from 13th to 28th inclusive) =

16/2 (78+246) = 1968

Please let me know if the MGMAT way is easier. If it is can u pls share it?

Thanks

Thanks a lot ri2007..this is exactly the type of solution I was looking for. Your solution is actually easier to me than MGMAT's solution. Nevertheless, here's their solution, thanks again!

"For sequence S, any value Sn equals 6n. Therefore, the problem can be restated as determining the sum of all multiples of 6 between 78 (S13) and 168 (S28), inclusive. The direct but time-consuming approach would be to manually add the terms: 78 + 84 = 162; 162 + 90 = 252; and so forth.

The solution can be found more efficiently by identifying the median of the set and multiplying by the number of terms. Because this set includes an even number of terms, the median equals the average of the two ‘middle’ terms, S20 and S21, or (120 + 126)/2 = 123. Given that there are 16 terms in the set, the answer is 16(123) = 1,968.

The correct answer is D. "

There is a formula:

S = (first + last)/2*number of terms

then S= (78+168)/2*16 = 1968

78 (S13) and 168 (S28),