Difference between revisions of "1962 AHSME Problems/Problem 3"
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==Problem== | ==Problem== | ||
− | The first three terms of an arithmetic progression are <math>x - 1, x + 1, 2x + 3</math>, in the order shown. The value of x is: | + | The first three terms of an arithmetic progression are <math>x - 1, x + 1, 2x + 3</math>, in the order shown. The value of <math>x</math> is: |
<math> \textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined} </math> | <math> \textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined} </math> | ||
==Solution== | ==Solution== | ||
− | + | Let <math>y</math> represent the common difference between the terms. We have <math>(x+1)-y=(x-1)\implies y=2</math>. | |
+ | |||
+ | Substituting gives us <math>(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0</math>. | ||
+ | |||
+ | Therefore, our answer is <math>\boxed{\textbf{(B)}\ 0}</math>; | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1962|before=Problem 2|num-a=4}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:12, 3 October 2014
Problem
The first three terms of an arithmetic progression are , in the order shown. The value of is:
Solution
Let represent the common difference between the terms. We have .
Substituting gives us .
Therefore, our answer is ;
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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All AHSME Problems and Solutions |
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