5lb book - Advanced Quant #54 (pg 1018)

[faedrus]

Apologies, I wasn't sure where to post this.

Can anyone recommend a technique for solving this problem? I'm not completely satisfied with the approach in the solutions as I haven't seen that particular approach applied to any of the other problems in this 1000+ page book.

Question: In a certain sequence, each term beyond the second term is equal to the average of the previous two terms. If a_1 and a_3 are positive integers, which of the following is NOT a possible value of a_5?

(A) -9/4
(B) 0
(C) 9/4
(D) 75/8
(E) 41/2

[faedrus]

I have a solution I'm happier with -- thought I'd share.

Let a_1 = x, a_3=y where x,y are integers.

Then, since a_3 is the average of a_1 and a_2, some simple algebra will get us a_2 = 2y - x. Applying the same formula for average, we get a_4 = (3y-x)/2 and a_5 = (5y-x)/4. Finally, since y and x are integers, we know that a_5 cannot have an 8 in the denominator, so the answer is D.

I like this solution more than focusing on the possible decimal representations of the solution.

[tommywallach]

Hey Faedrus,

I love your solution! I just want to add one more thing in case anybody else is confused on this.

75/8 = 9 and 3/8ths

Technically, you could have an unsimplified fraction over 8 that would work here. For example, 74/8 is secretly 37/4, so it works. But 75/8 is impossible, because any integer value divided by 4 will either be an integer, or end in .25, .5, or .75.

Great work!

-t