Three numbers form a geometric sequence, their sum is 13 , and the sum of they squares is 91. find these numbers.

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: \[ \boxed{1, 3, 9} \]