1972 AHSME Problems/Problem 16

Problem 16

There are two positive numbers that may be inserted between $3$ and $9$ such that the first three are in geometric progression while the last three are in arithmetic progression. The sum of those two positive numbers is

$\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }11\frac{1}{4}\qquad \textbf{(C) }10\frac{1}{2}\qquad \textbf{(D) }10\qquad \textbf{(E) }9\frac{1}{2}$


Let $a$ be first and $b$ be second. We can then get equations based on our knowledge: $b-a = 9-b$ and $b/a = a/3$. We then get $b = (9+a)/2$ from our first equation and we substitute that into the second to get $\frac{9+a}{2a} = a/3$ which simplifies to $2a^2-3a-27=0$ which be $(2a-9)(a+3) = 0$ and $a=9/2$. Then $b=27/4$. Their sums be $\boxed{\textbf{(B) }11\frac{1}{4}}$. ~lopkiloinm