2003 AMC 10B Problem 24

Question

The first four terms in an arithmetic sequence are

x+y,

x-y,

xy,

and

\frac{x}{y},

in that order. What is the fifth term?

-\dfrac{15}{8}

-\dfrac{6}{5}

0

\dfrac{27}{20}

\dfrac{123}{40}

The common difference is

(x-y)-(x+y)=-2y

So the third and fourth terms are

x-3y

and

x-5y;

set these equal to

xy

and

\frac{x}{y}

The common difference is

-2y,

so the third and fourth terms must be

x-3y

and

x-5y.

Thus

xy=x-3y

and

\dfrac{x}{y}=x-5y.

From

\dfrac{x}{y}=x-5y

we get

x=xy-5y^2,

and substituting

xy=x-3y

gives

-3y-5y^2=0.

Since

y\ne 0,

y=-\dfrac35

and then

x=-\dfrac98.

The fifth term is

x-7y=-\dfrac98 + \dfrac{21}{5}=\dfrac{123}{40}.

Thus, the correct answer is E.

Problem 24 in other years