# Difference between revisions of "1964 AHSME Problems/Problem 6"

## Problem

If $x, 2x+2, 3x+3, \dots$ are in geometric progression, the fourth term is:

$\textbf{(A)}\ -27 \qquad \textbf{(B)}\ -13\frac{1}{2} \qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\frac{1}{2}\qquad \textbf{(E)}\ 27$

## Solution

Since we know the sequence is a geometric sequence, the ratio of consecutive terms is always the same number. Thus, we can set up an equation:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$.

Solving it, we get:

$\frac{2x+2}{x} = \frac{3x+3}{2x+2}$

$(2x+2)(2x+2) = (3x+3)(x)$

$4x^2+8x+4 = 3x^2+3x$

$x^2+5x+4 = 0$

$(x + 4)(x+1) = 0$

$x = -4$ or $x = -1$

If $x=-1$, the sequence has a $0$ as the second term, which is not allowed in a geometric sequence, so it is an extraneous solution that came about because we cross-multiplied by $(2x + 2)$, which is $0$.

If $x=-4$, we plug into $x, 2x +2, 3x + 3$ to find the sequence starts as $-4, -6, -9$. The common ratio is $\frac{-6}{-4} = \frac{3}{2}$. The next term is $\frac{3}{2} \cdot -9 = \frac{-27}{2} = -13\frac{1}{2}$, which is option $\textbf{(B)}$