1978 AHSME Problems/Problem 8

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Problem 8

If $x\neq y$ and the sequences $x,a_1,a_2,y$ and $x,b_1,b_2,b_3,y$ each are in arithmetic progression, then $(a_2-a_1)/(b_2-b_1)$ equals

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


Solution

WLOG, let $x =2$ and $y = 5$. From the first sequence, we get \[2, 3, 4, 5\] so $a_2 - a_1 = 1$ From the second sequence, we get \[2, 2+r, 2+2r, 2+3r, 5\] so $2+4r = 5$ and $r = \frac{3}{4}$ Thus, we have \[2, 2 \frac{3}{4}, 3 \frac{1}{2} ...\] and $b_2 - b_1 = \frac{3}{4}$ So $\frac{1}{\frac{3}{4}} = \frac{4}{3}$ \[\boxed{D}\]

~JustinLee2017

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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