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 $\textbf{(B)}$

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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.

Page

Toolbox

Search

1964 AHSME Problems/Problem 6

Problem

Solution

See Also