8. The sum of the first \( n \) terms of a geometric progression is given by \( S_{n}=18\left[1-\left(\frac{1}{2}\right)^{n}\right] \) Find die conmun tatio of the proggession.

To find the common ratio of the geometric progression, we can use the formula for the sum of the first \( n \) terms of a geometric progression: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. Given the sum of the first \( n \) terms of the geometric progression is: \[ S_n = 18\left[1 - \left(\frac{1}{2}\right)^n\right] \] We can compare this with the formula for the sum of the first \( n \) terms of a geometric progression: \[ 18\left[1 - \left(\frac{1}{2}\right)^n\right] = \frac{a(1 - r^n)}{1 - r} \] Let's simplify and solve for \( r \): \[ 18\left[1 - \left(\frac{1}{2}\right)^n\right] = \frac{a(1 - r^n)}{1 - r} \] First, we can simplify the left side of the equation by distributing the 18: \[ 18 - 18\left(\frac{1}{2}\right)^n = \frac{a(1 - r^n)}{1 - r} \] Now, we can see that the term \( 18\left(\frac{1}{2}\right)^n \) is similar to the term \( a(1 - r^n) \) in the right side of the equation. Since the left side of the equation is a constant (18) minus a term that depends on \( n \), and the right side is a term that depends on \( n \), we can deduce that the common ratio \( r \) must be \( \frac{1}{2} \) because the term \( \left(\frac{1}{2}\right)^n \) in the left side corresponds to \( r^n \) in the right side. Therefore, the common ratio of the geometric progression is: \[ r = \frac{1}{2} \]