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

Revision as of 23:06, 9 November 2013

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

Revision as of 23:06, 9 November 2013 (view source) Fadebekun (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 23:06, 9 November 2013 (view source) Fadebekun (talk \| contribs) (→‎Solution) Newer edit →
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	Substituting gives us <math>(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0</math>.		Substituting gives us <math>(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0</math>.

−	Therefore, our answer is $\textbf{(B)}\ 0\qquad~~&#036~~;	+	Therefore, our answer is <math>\boxed{\textbf{(B)}\ 0\qquad}</math>;