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please solve it

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by tomasnmr » Fri Nov 23, 2012 10:55 pm
My solution:

If the sum (first 15) = sum (first 23), then sum (last 8) must be 0.

There simplest solution that makes an even series of numers 0 should be: -7, -5, -3, -1, 1, 3, 5, 7 Or -10, -7, but there are definitely other options aswell. Like -22, -17, -12, -7, -2, 2, 7, 12, 22.
However, none of the solutions will include 0 and they will all be symetric around 0. Note that 0 would be positioned between number 19 and number 20, and thus split the arithmetic progression in two. This means we have a sequence where all positive numbers are matched with a negative and the sum is 0.


IMO the answer is B. (edited, misread answer options)

First post here so sorry if it's a bit cluttered.

Bets of luck!
Last edited by tomasnmr on Sun Nov 25, 2012 12:41 am, edited 1 time in total.
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by GMATGuruNY » Sat Nov 24, 2012 1:52 pm
soumya_joy wrote:If the sum of the first 15 terms of an arithmetic progression equals that of the first 23
terms, then what is the sum of the first 38 terms?
A. -2 B. 0 C. 3 D. 5
Let the terms of the sequence be a�, a₂, a₃...
For the sum of the first 23 terms to be equal to the sum of the first 15 terms, the sum of terms a�₆ through a₂₃ must be 0.

For any arithmetic sequence, sum = (number of terms)(median).
Since a�₆ through a₂₃ is an arithmetic sequence composed of 8 terms whose sum is 0, we get:
0 = (8)(median)
median= 0.

a�₆ through a₂₃ are the 8 middle terms of a� through a₃₈.
Thus, the median of terms a�₆ through a₂₃ -- 0 -- is also the median of terms a� through a₃₈.
Thus:
Sum of the first 38 terms = (number)(median) = (38)(0) = 0.

The correct answer is B.
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