1978 AHSME Problems/Problem 8

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

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1978 AHSME Problems/Problem 8

Solution