1978 AHSME Problems/Problem 8: Difference between revisions
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== Problem 8 == | |||
If <math>x\neq y</math> and the sequences <math>x,a_1,a_2,y</math> and <math>x,b_1,b_2,b_3,y</math> each are in arithmetic progression, then <math>(a_2-a_1)/(b_2-b_1)</math> equals | |||
<math>\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} </math> | |||
==Solution== | ==Solution== | ||
WLOG, let <math>x =2</math> and <math>y = 5</math>. From the first sequence, we get | WLOG, let <math>x =2</math> and <math>y = 5</math>. From the first sequence, we get | ||
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~JustinLee2017 | ~JustinLee2017 | ||
==See Also== | |||
{{AHSME box|year=1978|num-b=7|num-a=9}} | |||
{{MAA Notice}} | |||
Latest revision as of 11:03, 13 February 2021
Problem 8
If 
and the sequences 
and 
each are in arithmetic progression, then 
equals
Solution
WLOG, let 
and 
. From the first sequence, we get
![\[2, 3, 4, 5\]](http://latex.artofproblemsolving.com/9/4/9/949877e06e49100198e46b1c662a9f0e46395fb5.png)
so 
From the second sequence, we get
![\[2, 2+r, 2+2r, 2+3r, 5\]](http://latex.artofproblemsolving.com/b/a/1/ba16a90006b31ed1d4e3140a6c987db94e5199ed.png)
so 
and 
Thus, we have ![\[2, 2 \frac{3}{4}, 3 \frac{1}{2} ...\]](http://latex.artofproblemsolving.com/a/6/4/a6480f242a146a65f56ea211cef3e59e03b24456.png)
and 
So 
![\[\boxed{D}\]](http://latex.artofproblemsolving.com/d/a/d/dadc8e71363a80c53a137421994b9e02de6eeabd.png)
~JustinLee2017
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
