1963 AHSME Problems/Problem 16

Problem

Three numbers $a,b,c$ , none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results in a geometric progression. Then $b$ equals:

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution

Let $d$ be the common difference of the arithmetic sequence, so $a = b-d$ and $c = b+d$ .

Since increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence, $\frac{b}{b-d+1} = \frac{b+d}{b}$ $\frac{b}{b-d} = \frac{b+d+2}{b}$ Cross-multiply in both equations to get a system of equations. $b^2 = b^2 - d^2 + b + d$ $b^2 = b^2 - d^2 + 2b - 2d$ Rearranging terms results in $d^2 = b+d$ $d^2 = 2b-2d$ Substitute and solve for $d$ . $b+d = 2b-2d$ $d = \frac{b}{3}$ Finally, substitute $d$ and solve for $b$ . Since $b \ne 0$ , dividing by $b$ is allowed. $(\frac{b}{3})^2 = b + \frac{b}{3}$ $\frac{b^2}{9} = \frac{4b}{3}$ $\frac{b}{9} = \frac{4}{3}$ $b = 12$ The answer is $\boxed{\textbf{(C)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1963 AHSME Problems/Problem 16

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