the distinct numbers x+9y, x+3y, x are the first 3 terms of a geometric sequence. find the common ration, and if the sum to infinity is 48, find x and y.

We are given that the numbers \( x + 9y \), \( x + 3y \), and \( x \) form the first three terms of a geometric sequence. Let's solve step by step.

### Step 1: The common ratio
For a geometric sequence, the ratio of consecutive terms is constant. Let the common ratio be \( r \). This gives us the following two equations:

\[
\frac{x + 3y}{x + 9y} = r \quad \text{(ratio between the 2nd and 1st term)}
\]
\[
\frac{x}{x + 3y} = r \quad \text{(ratio between the 3rd and 2nd term)}
\]

Since both expressions represent the common ratio \( r \), we can equate them:

\[
\frac{x + 3y}{x + 9y} = \frac{x}{x + 3y}
\]

### Step 2: Cross-multiply and simplify
Cross-multiply to eliminate the denominators:

\[
(x + 3y)(x + 3y) = x(x + 9y)
\]

Expanding both sides:

\[
(x + 3y)^2 = x(x + 9y)
\]
\[
x^2 + 6xy + 9y^2 = x^2 + 9xy
\]

Now, subtract \( x^2 \) from both sides:

\[
6xy + 9y^2 = 9xy
\]

Simplify by moving terms involving \( xy \) to one side:

\[
9y^2 = 3xy
\]

### Step 3: Solve for \( x \) in terms of \( y \)
We can now divide both sides by \( y \) (assuming \( y \neq 0 \)):

\[
9y = 3x
\]

This simplifies to:

\[
x = 3y
\]

### Step 4: Use the sum to infinity
The sum to infinity of a geometric series is given by:

\[
S_{\infty} = \frac{a}{1 - r}
\]

where \( a \) is the first term and \( r \) is the common ratio. We are given that the sum to infinity is 48, so:

\[
S_{\infty} = \frac{x + 9y}{1 - r} = 48
\]

From Step 2, we know that:

\[
r = \frac{x}{x + 3y}
\]

Substitute \( x = 3y \) into the expression for \( r \):

\[
r = \frac{3y}{3y + 3y} = \frac{3y}{6y} = \frac{1}{2}
\]

Now, substitute the value of \( r = \frac{1}{2} \) and \( x + 9y = 12y \) (since \( x = 3y \)) into the sum formula:

\[
\frac{12y}{1 - \frac{1}{2}} = 48
\]

Simplify:

\[
\frac{12y}{\frac{1}{2}} = 48
\]
\[
24y = 48
\]

Thus, \( y = 2 \).

### Step 5: Find \( x \)
Since \( x = 3y \), substitute \( y = 2 \):

\[
x = 3(2) = 6
\]

### Final Answer:
The values of \( x \) and \( y \) are:

\[
x = 6, \quad y = 2
\]

Would you like to see more details or have any questions?

Here are 5 related questions to expand the topic:
1. What is the general formula for the sum of an infinite geometric series?
2. How can you find the sum of the first \( n \) terms of a geometric sequence?
3. What conditions must be met for the sum to infinity to exist in a geometric series?
4. Can a geometric sequence have negative terms and still have a sum to infinity?
5. What are some real-life applications of geometric sequences and series?

**Tip:** Always check that the common ratio \( r \) is between -1 and 1 to ensure that the sum to infinity converges.

The distinct numbers x + 9y, x + 3y, x are the first 3 terms of a geometric sequence. Find the common ratio, and if the sum to infinity is 48, find x and y.

This problem involves finding the common ratio and solving for variables x and y in a geometric sequence. Learn how to use the sum to infinity formula and algebraic manipulation to solve this problem, suitable for Grades 9-11.

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