1978 AHSME Problems/Problem 8

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

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1978 AHSME Problems/Problem 8

Problem 8

Solution

See Also