AAPL wrote:Manhattan Prep

A certain series is defined by the following recursive rule: \(S_n=k(S_n-1)\), where \(k\) is a constant. If the 1st term of this series is 64 and the 25th term is 192, what is the 9th term?

A. \(\sqrt{2}\)
B. \(\sqrt{3}\)
C. \(64\sqrt{3}\)
D. \(64\cdot 3^{1/3}\)
E. \(64\cdot 3^{24}\)

OA D

This is not a "series". It is a "sequence". A series is a sum of a sequence, and this question has nothing to do with summing a sequence. It would be understandable for a test taker to be confused by the wording here (I was at first), because test takers aren't expected to know what a "series" even is.

There's also a typo in the question: the recursive rule should say S_n = k * S_(n-1). So we're multiplying one term in the sequence by k to find the next term in the sequence. Going from the first term to the 25th term, we'll multiply by k 24 times, so S_25 = k^24 * S_1. Since we know S_25 = 192 and S_1 = 64, we know 192 = k^24 * 64, so k^24 = 3, and k is the 24th root of 3, or 3^(1/24). Going from 64 = S_1 to S_9, we'll multiply by k eight times, so the answer is 64*(3^(1/24))^8 = 64*3^(1/3).