1987 AHSME Problems/Problem 9

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Problem

The first four terms of an arithmetic sequence are $a, x, b, 2x$. The ratio of $a$ to $b$ is

$\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3}\qquad \textbf{(E)}\ 2$

Solution

To get from the second term to the fourth term, we add $2$ lots of the common difference, so if the common difference is $d$, we have $2d = 2x -x = x \implies d = \frac{x}{2}$. Thus the sequence is $\frac{x}{2}, x, \frac{3x}{2}, 2x$, so the ratio is $\frac{\frac{x}{2}}{\frac{3x}{2}} = \frac{1}{3}$, which is answer $\boxed{\text{B}}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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