# Difference between revisions of "1950 AHSME Problems/Problem 5"

## Problem

If five geometric means are inserted between $8$ and $5832$, the fifth term in the geometric series:

$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$

## Solution

We can let the common ratio of the geometric sequence be $r$. $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$. By the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$ We divide by eight to find: $r^6=729$ $r=\pm 3$

Because $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\boxed{\textbf{(A)}\ 648}$