Bunuel wrote:

If p and n are positive integers, with p > n, what is the average of n consecutive multiples of p (starting with p), less the average of p consecutive multiples of n (starting with n)?

A. (p + n)/2

B. (p – n)/2

C. p + n

D. p – n

E. p^2 – n^2

MANHATTAN GMAT OFFICIAL SOLUTION:The easiest way to solve this problem is to pick convenient numbers for p and n. Say that p = 3 and n = 2, to follow the constraint that p > n.

Then we need to take the difference of two quantities:

1. The average of 2 consecutive multiples of 3 (starting with 3)

2. The average of 3 consecutive multiples of 2 (starting with 2).

The first quantity is the average of 3 and 6. That number is 4.5.

The second quantity is the average of 2, 4, and 6. That number is 4.

The difference between these two numbers is 0.5. We now look for the answer choice that yields 0.5 when p = 3 and n = 2. The only answer choice that does so is (p – n)/2.

Proving that this answer is correct in the abstract is significantly harder, but not outlandish.

The sum of n consecutive multiples of p = p + 2p + 3p + … + np.

Factor out a p to get p(1 + 2 + 3 + … + n). Now, the sum of n consecutive integers starting at 1 is the average of those integers times the number of integers.

The average of the integers is (n+ 1)/2. The number of integers is n.

So the sum of 1 through n is n(n + 1)/2. This is a formula that you shouldn’t memorize, but you should be distantly familiar with it, as passing acquaintances.

Thus, the sum of n consecutive multiples of p is pn(n + 1)/2, and the average of those n multiples is that sum divided by n, or p(n + 1)/2.

A similar argument gives you the average of p consecutive multiples of n. You can just switch p and n in the formula above: this average is n(p + 1)/2.

Subtracting these two gives you (p – n)/2.

The correct answer is B.
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