(1) In a convergent geometric series, the sum of the first four terms is \( \frac{80}{3} \) and the sum to infinity of the series is 27. Determine the general term.

To solve this problem, we start by recalling the formulas for the sum of a geometric series. The sum to infinity \( S \) for a convergent geometric series is given by \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio (with \( |r| < 1 \)). Here, we know: \[ S = 27 \implies \frac{a}{1 - r} = 27 \tag{1} \] Next, the sum of the first \( n \) terms of a geometric series is given by \[ S_n = a \frac{1 - r^n}{1 - r} \] We want the sum of the first four terms, \( S_4 \): \[ S_4 = a \frac{1 - r^4}{1 - r} = \frac{80}{3} \tag{2} \] Now we have two equations to work with: equation (1) for the sum to infinity and equation (2) for the sum of the first four terms. We can isolate \( a \) from (1): \[ a = 27(1 - r) \tag{3} \] Substituting (3) into (2): \[ 27(1 - r) \frac{1 - r^4}{1 - r} = \frac{80}{3} \] This simplifies to: \[ 27(1 - r^4) = \frac{80}{3} \] Now, multiply both sides by 3 to eliminate the fraction: \[ 81(1 - r^4) = 80 \] Then, divide by 81: \[ 1 - r^4 = \frac{80}{81} \] Thus, \[ r^4 = 1 - \frac{80}{81} = \frac{1}{81} \] Taking the fourth root gives us: \[ r = \frac{1}{3} \] Next, we substitute \( r \) back into equation (3) to find \( a \): \[ a = 27(1 - \frac{1}{3}) = 27 \times \frac{2}{3} = 18 \] Now we have \( a = 18 \) and \( r = \frac{1}{3} \). The general term of a geometric series is given by: \[ T_n = a \cdot r^{n-1} \] Substituting \( a \) and \( r \): \[ T_n = 18 \left(\frac{1}{3}\right)^{n-1} \] Thus, the general term of the given convergent geometric series is: \[ \boxed{18 \left(\frac{1}{3}\right)^{n-1}} \]