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Difference between revisions of "1978 AHSME Problems/Problem 8"

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== Problem 8 ==
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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
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<math>\textbf{(A) }\frac{2}{3}\qquad
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\textbf{(B) }\frac{3}{4}\qquad
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\textbf{(C) }1\qquad
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\textbf{(D) }\frac{4}{3}\qquad
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\textbf{(E) }\frac{3}{2} </math> 
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==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
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==See Also==
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{{AHSME box|year=1978|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 12:03, 13 February 2021

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 (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

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