# 1963 AHSME Problems/Problem 16

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