Question

If 3,x,12,y are in a geometric progression, find the product of x and y

Answer 1

In a geometric progression, the ratio of consecutive terms is constant. Given the terms \(3\), \(x\), \(12\), and \(y\), we can set up the relationships as follows:

The ratio between the first and second terms \( (x/3) \) must be equal to the ratio between the second and third terms \( (12/x) \). Therefore, we have:

\[
\frac{x}{3} = \frac{12}{x}
\]

Cross-multiplying gives:

\[
x^2 = 36
\]

Taking the square root of both sides leads to:

\[
x = 6 \quad \text{or} \quad x = -6
\]

Next, we continue with the second relationship between the third and fourth terms, \( (y/12) \) and \( (12/x) \):

\[
\frac{12}{x} = \frac{y}{12}
\]

Cross-multiplying gives:

\[
12^2 = xy
\]

This leads to:

\[
144 = xy
\]

Now, we can determine \(y\) for both potential values of \(x\):

1. If \(x = 6\):
\[
144 = 6y \implies y = \frac{144}{6} = 24
\]

2. If \(x = -6\):
\[
144 = (-6)y \implies y = \frac{144}{-6} = -24
\]

In both cases, we need the product of \(x\) and \(y\):

1. For \(x = 6\) and \(y = 24\):
\[
xy = 6 \times 24 = 144
\]

2. For \(x = -6\) and \(y = -24\):
\[
xy = (-6) \times (-24) = 144
\]

Thus, the product \(xy\) is consistent in both cases.

Finally, the product of \(x\) and \(y\) is:

\[
\boxed{144}
\]