(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 find the general term of a convergent geometric series, we can use the formula for the sum of the first n terms of a geometric series: \[ S_n = \frac{a(1 - r^n)}{1 - r} \] where: - \( S_n \) is the sum of the first n terms, - \( a \) is the first term, - \( r \) is the common ratio. Given that the sum of the first four terms is \( \frac{80}{3} \) and the sum to infinity is 27, we can set up the following equations: 1. For the sum of the first four terms: \[ S_4 = \frac{a(1 - r^4)}{1 - r} = \frac{80}{3} \] 2. For the sum to infinity: \[ S_{\infty} = \frac{a}{1 - r} = 27 \] We can solve these equations simultaneously to find the values of \( a \) and \( r \), and then use them to determine the general term of the series. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{a\left(1-r^{4}\right)}{1-r}=\frac{80}{3}\\\frac{a}{1-r}=27\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}a\left(1+r+r^{2}+r^{3}\right)=\frac{80}{3}\\\frac{a}{1-r}=27\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}a=\frac{80}{3+3r+3r^{2}+3r^{3}}\\\frac{a}{1-r}=27\end{array}\right.\) - step3: Substitute the value of \(a:\) \(\frac{\frac{80}{3+3r+3r^{2}+3r^{3}}}{1-r}=27\) - step4: Simplify: \(\frac{80}{\left(3+3r+3r^{2}+3r^{3}\right)\left(1-r\right)}=27\) - step5: Cross multiply: \(80=\left(3+3r+3r^{2}+3r^{3}\right)\left(1-r\right)\times 27\) - step6: Simplify the equation: \(80=27\left(3+3r+3r^{2}+3r^{3}\right)\left(1-r\right)\) - step7: Swap the sides: \(27\left(3+3r+3r^{2}+3r^{3}\right)\left(1-r\right)=80\) - step8: Expand the expression: \(81-81r^{4}=80\) - step9: Move the constant to the right side: \(-81r^{4}=80-81\) - step10: Subtract the numbers: \(-81r^{4}=-1\) - step11: Change the signs: \(81r^{4}=1\) - step12: Divide both sides: \(\frac{81r^{4}}{81}=\frac{1}{81}\) - step13: Divide the numbers: \(r^{4}=\frac{1}{81}\) - step14: Simplify the expression: \(r=\pm \sqrt[4]{\frac{1}{81}}\) - step15: Simplify: \(r=\pm \frac{1}{3}\) - step16: Separate into possible cases: \(r=\frac{1}{3}\cup r=-\frac{1}{3}\) - step17: Rearrange the terms: \(\left\{ \begin{array}{l}a=\frac{80}{3+3r+3r^{2}+3r^{3}}\\r=\frac{1}{3}\end{array}\right.\cup \left\{ \begin{array}{l}a=\frac{80}{3+3r+3r^{2}+3r^{3}}\\r=-\frac{1}{3}\end{array}\right.\) - step18: Calculate: \(\left\{ \begin{array}{l}a=18\\r=\frac{1}{3}\end{array}\right.\cup \left\{ \begin{array}{l}a=36\\r=-\frac{1}{3}\end{array}\right.\) - step19: Check the solution: \(\left\{ \begin{array}{l}a=18\\r=\frac{1}{3}\end{array}\right.\cup \left\{ \begin{array}{l}a=36\\r=-\frac{1}{3}\end{array}\right.\) - step20: Rewrite: \(\left(a,r\right) = \left(18,\frac{1}{3}\right)\cup \left(a,r\right) = \left(36,-\frac{1}{3}\right)\) The solutions to the system of equations are: 1. \( a = 18 \) and \( r = \frac{1}{3} \) 2. \( a = 36 \) and \( r = -\frac{1}{3} \) Since the series is convergent, the common ratio \( r \) must be between -1 and 1. Therefore, the correct solution is \( a = 18 \) and \( r = \frac{1}{3} \). Now, we can use the formula for the general term of a geometric series: \[ a_n = a \cdot r^{n-1} \] Substitute the values of \( a \) and \( r \) into the formula to find the general term of the series. Simplify the expression by following steps: - step0: Solution: \(18\left(\frac{1}{3}\right)^{n-1}\) - step1: Multiply the terms: \(18\times 3^{-n+1}\) The general term of the convergent geometric series is \( 18 \times 3^{-n+1} \).