7 The sum to \( n \) terms of a geometric series is 728 . If the first term is 2 and the last term is 486 , find the common ratio and the number of terms.

To solve this problem, we start with the known formulas for a geometric series. The sum \( S_n \) of the first \( n \) terms of a geometric series can be represented as: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In this case, we have \( S_n = 728 \) and \( a = 2 \). We also know the last term (\( a_n \)) of the geometric series can be expressed as: \[ a_n = a r^{n-1} \] Given that \( a_n = 486 \), we can substitute the known values: \[ 2 r^{n-1} = 486 \] From this, we can solve for \( r^{n-1} \): \[ r^{n-1} = \frac{486}{2} = 243 \] Now we have two equations to work with: 1. \( 728 = 2 \frac{1 - r^n}{1 - r} \) 2. \( r^{n-1} = 243 \) We'll start by calculating \( r \) from the second equation. Since \( 243 = 3^5 \), we can express \( r^{n-1} \) as: \[ r^{n-1} = 3^5 \] Let’s assume \( r = 3 \), then: \[ 3^{n-1} = 3^5 \implies n-1 = 5 \implies n = 6 \] Now, substituting \( r = 3 \) and \( n = 6 \) into the first equation to check if both relationships hold: \[ S_n = 2 \frac{1 - 3^6}{1 - 3} = 2 \frac{1 - 729}{-2} = 2 \cdot \frac{-728}{-2} = 728 \] Both conditions are satisfied, confirming that \( r = 3 \) and \( n = 6 \). Thus, the common ratio is \( 3 \) and the number of terms is \( 6 \).