Bunuel wrote:

In a certain sequence, the term \(A_n\) is given by the formula \(A_n= 16-\frac{(4- n)^3}{|n-4|}\) for all positive integers n ≠ 4, while \(A_4 = 16\). For which integer value of k greater than 2 is the mean of all values in the sequence from \(A_1\) through \(A_k\) equal to the median of those values?

(A) 6

(B) 7

(C) 8

(D) 9

(E) 10

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:As is the case with most sequence problems, your first step is to compute the first several terms. The formula looks ugly, but if you work from the inside out, respecting order of operations, you’ll be in good shape. Do the calculations carefully and clearly, not only so that you don’t make a mistake but also so that you can spot emerging patterns.

\(A_n=16-\frac{(4-n)^3}{|n – 4|}\)

If n = 1, then \(A_1=16-\frac{(4-1)^3}{|1 – 4|}\)

\(= 16-\frac{(3)^3}{|- 3|}\)

\(= 16-\frac{3^3}{3}\)

\(= 16-3^2\)

\(= 16 – 9\)

\(= 7\)

If n = 2, then \(A_2=16-\frac{(4-2)^3}{|2 – 4|}\)

\(= 16-\frac{(2)^3}{|- 2|}\)

\(= 16-\frac{2^3}{2}\)

\(= 16-2^2\)

\(= 16 – 4\)

\(= 12\)

If n = 3, then \(A_3=16-\frac{(4-3)^3}{|3 – 4|}\)

\(= 16-\frac{(1)^3}{|- 1|}\)

\(= 16-\frac{1^3}{1}\)

\(= 16-1^2\)

\(= 16 – 1\)

\(= 15\)

So far, the sequence seems to be converging on 16. The first term, \(A_1\), is 16 minus 9; the second term is 16 minus 4; the third term is 16 minus 1. The separate definition of \(A_4\) as 16 confirms our suspicion. (The purpose of the separate definition is to prevent division by 0 in that one case.)

What happens when n = 5?

If n = 5, then \(A_5=16-\frac{(4-5)^3}{|5 – 4|}\)

\(= 16-\frac{(-1^3)}{|1|}\)

\(= 16+\frac{1^3}{1}\)

\(= 16+1^2\)

\(= 16 + 1\)

\(= 17\)

So, on the other side of 16, the terms pick up by adding a square to 16.

Consider what we are asked. We are looking for how far we need to go past the second term to make the mean of all the values up to that point equal the median. For the mean to equal the median, the values must be symmetrically spaced around the mean. Our sequence is not evenly spaced, but at a certain point, it will be symmetrically spaced around some number.

From the special role of 16, we might suspect that 16 will wind up being the number around which the sequence will be symmetrically spaced. That suspicion is correct:

If n = 1, then \(A_1=16-3^2=16-9 =7\)

If n = 2, then \(A_2=16-2^2=16-4=12\)

If n = 3, then \(A_3=16-1^2 =16-1=15\)

If n = 4, then \(A_4=16\)

If n = 5, then \(A_5=16+1^2=16+1=17\)

If n = 6, then \(A_6=16+2^2=16+4=20\)

If n = 7, then \(A_7=16+3^2=16+9=25\)

At that point, the sequence is symmetrical around 16. For all higher values of n, the sequence is no longer symmetrical, so the mean and the median will not be equal.

The correct answer is B.
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