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1962 AHSME Problems/Problem 3

Revision as of 00:06, 10 November 2013 by Fadebekun (talk | contribs) (→‎Solution)

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

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