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

## 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 (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS