Pregunta
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The sum of the first \( n \) terms of a geometric series is denoted by \( S_{n} \). The common ratio of the series, \( r \), is greater than 1. a) If \( S_{4}=5 S_{2} \) find the value of \( r \). b) Given further that \( S_{3}=21 \) determine the value of \( S_{10} \).

Ask by Stephens Williams. in Ghana
Feb 15,2025

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a) \( r = 2 \) b) \( S_{10} = 3069 \)

Solución

We are given a geometric series with first term a and common ratio r, with r > 1. The sum of the first n terms is given by   Sₙ = a(rⁿ – 1)/(r – 1). Part (a): We're given that S₄ = 5S₂.   • Write S₄ = a(r⁴ – 1)/(r – 1) and S₂ = a(r² – 1)/(r – 1).   • Substitute into the equation:     a(r⁴ – 1)/(r – 1) = 5·[a(r² – 1)/(r – 1)]     Cancel a and r – 1 (both nonzero):     r⁴ – 1 = 5(r² – 1)   • Expand and simplify:     r⁴ – 1 = 5r² – 5     r⁴ – 5r² + 4 = 0   • Factor this as a quadratic in r². Let x = r²:     x² – 5x + 4 = 0  ⟹ (x – 1)(x – 4) = 0.   • Thus, x = 1 or x = 4. Since x = r² and r > 1, we choose:     r² = 4  ⟹ r = 2. Part (b): We are given that S₃ = 21.   • Using r = 2:     S₃ = a(2³ – 1)/(2 – 1) = a(8 – 1)/1 = 7a.     Set 7a = 21 ⟹ a = 3.   • Now, find S₁₀:     S₁₀ = 3(2¹⁰ – 1)/(2 – 1) = 3(1024 – 1) = 3(1023) = 3069. Hence, the answers are:   a) r = 2   b) S₁₀ = 3069.

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Bonus Knowledge

To solve the problem, we will break it down step by step. ### a) Finding the value of \( r \) The sum of the first \( n \) terms of a geometric series can be represented using the formula: \[ S_{n} = a \frac{r^n - 1}{r - 1} \] where \( a \) is the first term and \( r \) is the common ratio. Given that \( S_{4} = 5 S_{2} \): 1. Calculate \( S_{2} \): \[ S_{2} = a \frac{r^2 - 1}{r - 1} \] 2. Calculate \( S_{4} \): \[ S_{4} = a \frac{r^4 - 1}{r - 1} \] 3. Substitute these into the equation \( S_{4} = 5 S_{2} \): \[ a \frac{r^4 - 1}{r - 1} = 5 \left( a \frac{r^2 - 1}{r - 1} \right) \] 4. The \( a \) and \( (r - 1) \) terms can be canceled (assuming \( a \neq 0 \) and \( r \neq 1 \)): \[ \frac{r^4 - 1}{r - 1} = 5 \frac{r^2 - 1}{r - 1} \] 5. Multiply both sides by \( (r - 1) \): \[ r^4 - 1 = 5(r^2 - 1) \] 6. Expand and rearrange the equation: \[ r^4 - 5r^2 + 4 = 0 \] 7. Let \( x = r^2 \). The equation becomes: \[ x^2 - 5x + 4 = 0 \] 8. Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \] This gives us: \[ x = 4 \quad \text{or} \quad x = 1 \] 9. Since \( x = r^2 \), we get: \[ r^2 = 4 \implies r = 2 \quad (\text{since } r > 1) \] Thus, the common ratio \( r \) is \( 2 \). ### b) Finding the value of \( S_{10} \) given \( S_{3} = 21 \) Now that we know \( r = 2 \), we can find \( a \) using the equation for \( S_{3} \). 1. We know: \[ S_{3} = a \frac{r^3 - 1}{r - 1} = 21 \] Substituting \( r = 2 \): \[ S_{3} = a \frac{2^3 - 1}{2 - 1} = a \frac{8 - 1}{1} = 7a \] Equating this to 21 gives: \[ 7a = 21 \implies a = 3 \] 2. Now we need to find \( S_{10} \): \[ S_{10} = a \frac{r^{10} - 1}{r - 1} \] Substituting \( a = 3 \) and \( r = 2 \): \[ S_{10} = 3 \frac{2^{10} - 1}{2 - 1} = 3(1024 - 1) = 3 \times 1023 = 3069 \] ### Final Answers a) The value of \( r \) is \( 2 \). b) The value of \( S_{10} \) is \( 3069 \).

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