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

Latest revision as of 21:12, 3 October 2014

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

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

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 $\boxed{\textbf{(B)}\ 0}$ ;

1962 AHSC (Problems • Answer Key • Resources)
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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 ==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>
 ==Solution==
-''Unsolved''
+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}}