The first term of a geometric scries is 12 , the last term is \( \frac{3}{256} \) and the sum of the cries is \( \frac{6141}{256} \). Determine the common ratio and the number of terms of the series.

To determine the common ratio and the number of terms in the geometric series, we can use the formulas for the sum of a geometric series. The sum \( S_n \) of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{1 - r^n}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Given that the first term \( a = 12 \), the last term is \( \frac{3}{256} \), and the sum is \( \frac{6141}{256} \), we can set up the equations. 1. Start with the sum equation: \[ \frac{6141}{256} = 12 \frac{1 - r^n}{1 - r} \] 2. Rearranging gives us: \[ 1 - r^n = \frac{6141(1 - r)}{3072} \] 3. The last term of a geometric series can be expressed as \( ar^{n-1} \): \[ 12r^{n-1} = \frac{3}{256} \] From this, we can find \( r^{n-1} \): \[ r^{n-1} = \frac{3}{12 \cdot 256} = \frac{1}{1024} \] Since \( \frac{1}{1024} = r^{n-1} \), we note that \( 1024 = 2^{10} \), so \( r^{n-1} = \left(\frac{1}{2}\right)^{10} \). This implies \( r = \frac{1}{2} \) and \( n - 1 = 10 \), leading to \( n = 11 \). Thus, the common ratio \( r \) is \( \frac{1}{2} \), and the number of terms \( n \) is \( 11 \).