Answer
The three numbers are 1, 3, and 9.
Solution
Let's denote the three numbers in the geometric sequence as \( a \), \( ar \), and \( ar^2 \), where \( a \) is the first term and \( r \) is the common ratio.
### Step 1: Set up the equations
1. The sum of the numbers:
\[
a + ar + ar^2 = 13
\]
This can be factored as:
\[
a(1 + r + r^2) = 13 \quad \text{(Equation 1)}
\]
2. The sum of the squares of the numbers:
\[
a^2 + (ar)^2 + (ar^2)^2 = 91
\]
This can be rewritten as:
\[
a^2(1 + r^2 + r^4) = 91 \quad \text{(Equation 2)}
\]
### Step 2: Solve the equations
From Equation 1, we can express \( a \) in terms of \( r \):
\[
a = \frac{13}{1 + r + r^2}
\]
Now, substitute this expression for \( a \) into Equation 2:
\[
\left(\frac{13}{1 + r + r^2}\right)^2(1 + r^2 + r^4) = 91
\]
### Step 3: Simplify and solve for \( r \)
Let's simplify this equation:
\[
\frac{169(1 + r^2 + r^4)}{(1 + r + r^2)^2} = 91
\]
Cross-multiplying gives:
\[
169(1 + r^2 + r^4) = 91(1 + r + r^2)^2
\]
Now, we will expand both sides and simplify.
1. Left side:
\[
169 + 169r^2 + 169r^4
\]
2. Right side:
\[
91(1 + 2r + r^2) = 91 + 182r + 91r^2
\]
Expanding gives:
\[
91 + 182r + 91r^2
\]
Now, we can set the two sides equal:
\[
169 + 169r^2 + 169r^4 = 91 + 182r + 91r^2
\]
### Step 4: Rearranging the equation
Rearranging gives:
\[
169r^4 + (169 - 91)r^2 - 182r + (169 - 91) = 0
\]
\[
169r^4 + 78r^2 - 182r + 78 = 0
\]
### Step 5: Solve the quartic equation
Now, we will solve this quartic equation for \( r \).
Let's use a numerical method or a tool to find the roots of this polynomial.
I'll proceed to solve this quartic equation.
Solve the equation(The complex numbers system) by following steps:
- step0: Solve for \(r\) in the complex numbers system:
\(169r^{4}+78r^{2}-182r+78=0\)
- step1: Factor the expression:
\(13\left(13r^{4}+6r^{2}-14r+6\right)=0\)
- step2: Divide both sides:
\(13r^{4}+6r^{2}-14r+6=0\)
- step3: Calculate:
\(\begin{align}&r\approx 0.550582+0.211935i\\&r\approx -0.550582+1.011387i\\&r\approx 0.550582-0.211935i\\&r\approx -0.550582-1.011387i\end{align}\)
- step4: Rewrite:
\(r_{1}\approx -0.550582-1.011387i,r_{2}\approx 0.550582-0.211935i,r_{3}\approx -0.550582+1.011387i,r_{4}\approx 0.550582+0.211935i\)
- step5: Remove the complex number(s):
\(\textrm{No real solution}\)
It appears that the quartic equation \( 169r^4 + 78r^2 - 182r + 78 = 0 \) has no real solutions. This suggests that the numbers we are looking for may not exist in the real number system under the given conditions.
### Step 6: Re-evaluate the problem
Let's re-examine the conditions:
1. The sum of the numbers is 13.
2. The sum of their squares is 91.
Given that the geometric sequence is defined as \( a \), \( ar \), and \( ar^2 \), we can also try to express the conditions differently or check for possible integer solutions.
### Step 7: Check for integer solutions
Since the numbers are in a geometric sequence, we can assume they might be integers. Let's denote the three numbers as \( x \), \( y \), and \( z \) such that:
- \( x + y + z = 13 \)
- \( x^2 + y^2 + z^2 = 91 \)
We can also express \( y \) and \( z \) in terms of \( x \) and a common ratio \( r \):
- \( y = xr \)
- \( z = xr^2 \)
Let's try some integer values for \( r \) and see if we can find a valid combination.
### Step 8: Testing integer values
1. If \( r = 1 \):
- The numbers would be \( x, x, x \) which gives \( 3x = 13 \) (not an integer).
2. If \( r = 2 \):
- The numbers would be \( x, 2x, 4x \).
- The sum: \( x + 2x + 4x = 7x = 13 \) → \( x = \frac{13}{7} \) (not an integer).
3. If \( r = \frac{1}{2} \):
- The numbers would be \( x, \frac{x}{2}, \frac{x}{4} \).
- The sum: \( x + \frac{x}{2} + \frac{x}{4} = \frac{7x}{4} = 13 \) → \( x = \frac{52}{7} \) (not an integer).
4. If \( r = 3 \):
- The numbers would be \( x, 3x, 9x \).
- The sum: \( x + 3x + 9x = 13x = 13 \) → \( x = 1 \).
- The numbers are \( 1, 3, 9 \).
- Check the sum of squares: \( 1^2 + 3^2 + 9^2 = 1 + 9 + 81 = 91 \) (valid).
### Conclusion
The three numbers that form a geometric sequence, sum to 13, and have a sum of squares equal to 91 are:
\[
\boxed{1, 3, 9}
\]
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