11. Prove that the product of \( n \) geometric means between \( a \) and \( b \) to the nth power of the single geometric mean between them. For what value of \( n, \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} \) is the geometric mean betweo iw?

Certainly! Let's tackle the problem step by step. ### **Problem Statement** 1. **Proof**: Show that the product of \( n \) geometric means between \( a \) and \( b \) is equal to the \( n \)th power of the single geometric mean between them. 2. **Find \( n \)**: Determine the value of \( n \) for which \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \] equals the geometric mean between \( a \) and \( b \). --- ### **1. Proving the Product of \( n \) Geometric Means** **Understanding Geometric Means:** Suppose there are \( n \) geometric means between \( a \) and \( b \). This sequence forms a geometric progression: \[ a, G_1, G_2, \dots, G_n, b \] Here, \( G_1, G_2, \dots, G_n \) are the geometric means, and the common ratio is \( r \). **Expressing \( b \) in Terms of \( a \) and \( r \):** \[ b = a \cdot r^{n+1} \] \[ \Rightarrow r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \] **Product of the \( n \) Geometric Means:** \[ G_1 \times G_2 \times \dots \times G_n = a^n \cdot r^{1 + 2 + \dots + n} = a^n \cdot r^{\frac{n(n+1)}{2}} \] Substituting \( r \): \[ = a^n \cdot \left(\frac{b}{a}\right)^{\frac{n}{2}} = (a^{\frac{n}{2}} \cdot b^{\frac{n}{2}}) = (ab)^{\frac{n}{2}} \] **Single Geometric Mean:** \[ \sqrt{ab} \] **Raising to the \( n \)th Power:** \[ (\sqrt{ab})^n = (ab)^{\frac{n}{2}} \] **Conclusion:** \[ G_1 \times G_2 \times \dots \times G_n = (\sqrt{ab})^n \] Thus, the product of \( n \) geometric means between \( a \) and \( b \) is indeed equal to the \( n \)th power of the single geometric mean between them. --- ### **2. Finding the Value of \( n \)** We need to find \( n \) such that: \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \sqrt{ab} \] **Let’s Introduce a Substitution:** Let \( k = \frac{a}{b} \). Assume \( b \neq 0 \) and \( k \neq 1 \) (since if \( k = 1 \), \( a = b \), and the equation trivially holds for any \( n \)). **Rewrite the Equation in Terms of \( k \):** \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = b \cdot \frac{k^{n+1} + 1}{k^n + 1} = \sqrt{ab} = b \sqrt{k} \] Divide both sides by \( b \): \[ \frac{k^{n+1} + 1}{k^n + 1} = \sqrt{k} \] **Cross-Multiplying:** \[ k^{n+1} + 1 = \sqrt{k} \cdot (k^n + 1) \] \[ k^{n+1} + 1 = k^{n + \frac{1}{2}} + \sqrt{k} \] Rearranging: \[ k^{n+1} - k^{n + \frac{1}{2}} + 1 - \sqrt{k} = 0 \] Factor by grouping: \[ k^{n + \frac{1}{2}}(k^{\frac{1}{2}} - 1) + (1 - k^{\frac{1}{2}}) = 0 \] \[ (k^{\frac{1}{2}} - 1)(k^{n + \frac{1}{2}} - 1) = 0 \] Given that \( k \neq 1 \), the only solution comes from: \[ k^{n + \frac{1}{2}} = 1 \] Since \( k > 0 \) and \( k \neq 1 \), the exponent must be zero: \[ n + \frac{1}{2} = 0 \] \[ n = -\frac{1}{2} \] **Conclusion:** The value of \( n \) that satisfies the given equation is \( n = -\frac{1}{2} \). --- ### **Final Answer** 1. **Proof**: The product of \( n \) geometric means between \( a \) and \( b \) is equal to \( (\sqrt{ab})^n \). 2. **Value of \( n \)**: The value of \( n \) for which \[ \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \] equals the geometric mean between \( a \) and \( b \) is \( n = -\frac{1}{2} \). **Answer**: After simplifying, the required value is –½. Thus, 1. The product of n geometric means between a and b is (√(ab))ⁿ. 2. When n = −½, (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) is equal to the geometric mean of a and b.