Difference between revisions of "1962 AHSME Problems/Problem 3"

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==Solution==
 
==Solution==
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Let <math>y</math> represent the common difference between the terms. We have <math>(x+1)-y=(x-1)\implies y=2</math>.
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Substituting gives us <math>(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0</math>.
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Therefore, our answer is <math>\boxed{\textbf{(B)}\ 0}</math>;
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==See Also==
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{{AHSME 40p box|year=1962|before=Problem 2|num-a=4}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:12, 3 October 2014

Problem

The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:

$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$

Solution

Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\implies y=2$.

Substituting gives us $(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0$.

Therefore, our answer is $\boxed{\textbf{(B)}\ 0}$;

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

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