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

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

Solution 1

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 sum is $\boxed{\textbf{(B) }11\frac{1}{4}}$. ~lopkiloinm

Solution 2

Like in Solution 1, let $a$ be the first number and $b$ be the second. From the problem, we have that

$a = 3r$ $\textbf{(1)}$

$b = 3r^2$ $\textbf{(2)}$

$9 = a + 2d$ $\textbf{(3)}$

for some common difference $d$ and common ratio $r$. We can use these to obtain the following:

$3r + d = 3r^2$ (from $\textbf{(1)}$, $\textbf{(2)}$, and the definition of an arithmetic sequence) $\textbf{(4)}$

$9 = 3r + 2d$ (from $\textbf{(1)}$ and $\textbf{(3)}$) $\textbf{(5)}$

Solving $\textbf{(5)}$ for $d$ and substituting into $\textbf{(4)}$, we find that $3r + \frac{9 - 3r}{2} = 3r^2$ or $6r + (9 - 3r) = 6r^2$. Moving every term to the right side and factoring, we get that $3(2r - 3)(r + 1) = 0$. Because $a$ and $b$ must be positive, the common ratio cannot be $-1$, so we ascertain that the common ratio is $3/2$. From here, we find that $a = 3 \cdot \frac{3}{2} = \frac{9}{2}$, and $b = \frac{9}{2} \cdot \frac{3}{2} = 27/4$. Then their sum is $\boxed{\textbf{(B) }11\frac{1}{4}}$.

~ cxsmi