The first term of a geometric sequence is 3 and the sum of the second and third term is 60 . Determine the first three terms and the general term Th.

Let's determine the first three terms and the general term \( T_h \) of the geometric sequence. ### Given: - **First term (\( a \))**: 3 - **Sum of the second and third terms**: 60 ### Steps to Find the Common Ratio (\( r \)): 1. **Express the Second and Third Terms in Terms of \( r \):** - **Second term**: \( a \cdot r = 3r \) - **Third term**: \( a \cdot r^2 = 3r^2 \) 2. **Set Up the Equation Based on the Given Sum:** \[ 3r + 3r^2 = 60 \] 3. **Simplify the Equation:** \[ r + r^2 = 20 \quad \Rightarrow \quad r^2 + r - 20 = 0 \] 4. **Solve the Quadratic Equation for \( r \):** \[ r = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2} \] - **First solution**: \[ r = \frac{-1 + 9}{2} = 4 \] - **Second solution**: \[ r = \frac{-1 - 9}{2} = -5 \] ### Determining the Sequences: 1. **When \( r = 4 \):** - **First term**: 3 - **Second term**: \( 3 \times 4 = 12 \) - **Third term**: \( 3 \times 4^2 = 48 \) - **General term (\( T_h \))**: \[ T_h = 3 \times 4^{h-1} \] 2. **When \( r = -5 \):** - **First term**: 3 - **Second term**: \( 3 \times (-5) = -15 \) - **Third term**: \( 3 \times (-5)^2 = 75 \) - **General term (\( T_h \))**: \[ T_h = 3 \times (-5)^{h-1} \] ### **Final Answer:** There are two possible geometric sequences: 1. **With common ratio \( r = 4 \):** - **First three terms**: 3, 12, 48 - **General term**: \( T_h = 3 \times 4^{\,h-1} \) 2. **With common ratio \( r = -5 \):** - **First three terms**: 3, –15, 75 - **General term**: \( T_h = 3 \times (-5)^{\,h-1} \)