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 \( S_{10} \)

To solve part (a), we need to use the formula for the sum of the first \( n \) terms of a geometric series: \[ S_{n} = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Given \( S_{4} = 5 S_{2} \), we can express both sums as follows: \[ S_{4} = a \frac{1 - r^4}{1 - r} \] \[ S_{2} = a \frac{1 - r^2}{1 - r} \] Substituting these into the equation \( S_{4} = 5 S_{2} \): \[ a \frac{1 - r^4}{1 - r} = 5 \left( a \frac{1 - r^2}{1 - r} \right) \] Assuming \( a \neq 0 \) and simplifying, we get: \[ 1 - r^4 = 5(1 - r^2) \] Expanding this yields: \[ 1 - r^4 = 5 - 5r^2 \] Rearranging gives: \[ r^4 - 5r^2 + 4 = 0 \] Let \( x = r^2 \), so the equation becomes: \[ x^2 - 5x + 4 = 0 \] Factoring the quadratic, we get: \[ (x - 1)(x - 4) = 0 \] Thus, \( x = 1 \) or \( x = 4 \). Converting back to \( r \): - If \( r^2 = 1 \), then \( r = 1 \) (not valid since \( r > 1 \)). - If \( r^2 = 4 \), then \( r = 2 \). So, the value of \( r \) is \( r = 2 \). For part (b), since we know \( S_{3} = 21 \) and we found \( r = 2 \), we can also express \( S_{3} \): \[ S_{3} = a \frac{1 - r^3}{1 - r} = a \frac{1 - 2^3}{1 - 2} = a \frac{1 - 8}{-1} = 7a \] Setting \( S_{3} \) equal to 21 gives us: \[ 7a = 21 \implies a = 3 \] Now we find \( S_{10} \): \[ S_{10} = a \frac{1 - r^{10}}{1 - r} = 3 \frac{1 - 2^{10}}{1 - 2} \] Calculating \( 2^{10} = 1024 \): \[ S_{10} = 3 \frac{1 - 1024}{-1} = 3 \times 1023 = 3069 \] Thus, \( S_{10} = 3069 \). Final answers: a) The value of \( r \) is \( 2 \). b) The value of \( S_{10} \) is \( 3069 \).