MathDB
Find (a_2-a_1)/(b_2-b_1)

Source: 1978 AHSME Problem 8

May 31, 2014
arithmetic sequenceAMC

Problem Statement

If x≠yx\neq y and the sequences x,a1,a2,yx,a_1,a_2,y and x,b1,b2,b3,yx,b_1,b_2,b_3,y each are in arithmetic progression, then (a2−a1)/(b2−b1)(a_2-a_1)/(b_2-b_1) equals
<spanclass=′latex−bold′>(A)</span>23<spanclass=′latex−bold′>(B)</span>34<spanclass=′latex−bold′>(C)</span>1<spanclass=′latex−bold′>(D)</span>43<spanclass=′latex−bold′>(E)</span>32<span class='latex-bold'>(A) </span>\frac{2}{3}\qquad<span class='latex-bold'>(B) </span>\frac{3}{4}\qquad<span class='latex-bold'>(C) </span>1\qquad<span class='latex-bold'>(D) </span>\frac{4}{3}\qquad <span class='latex-bold'>(E) </span>\frac{3}{2}
Back to ProblemsView on AoPS