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 $\boxed{\textbf{(C)}}$.


See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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