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

Solution 2

Like in Solution 1, let $a$ be the first number and $b$ be the second. From the problem, we have that $a = 3r$ and $b = 3r^2$ for some common ratio r. Also, $9 = a + 2d$ for some common difference d. We use the first and second listed equations to get that $3r + d = 3r^2$ , and we use the first and third equations to find that $9 = 3r + 2d$ . Solving the latter equation for d and substituting into the former, we find that $3r + \frac{9 - 3r}{2} = 3r^2$ or $6r + (9 - 3r) = 6r^2$ . Moving every term to one side and factoring, we get that $3(2r - 3)(r + 1) = 0$ . Clearly, 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

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1972 AHSME Problems/Problem 16

Problem 16

Solution 1

Solution 2