The first term of a geometric sequence is 3 and the sum of the second and third terms is 60. Determine the first three terms and the general term \( T_{n} \).

To find the terms of the geometric sequence, we denote the common ratio as \( r \). The first term \( T_1 = 3 \), the second term \( T_2 = 3r \), and the third term \( T_3 = 3r^2 \). According to the problem, the sum of the second and third terms is 60: \[ T_2 + T_3 = 3r + 3r^2 = 60 \] This simplifies to: \[ 3r(1 + r) = 60 \] Dividing by 3 gives: \[ r(1 + r) = 20 \] Rearranging leads to the quadratic equation: \[ r^2 + r - 20 = 0 \] Factoring, we find: \[ (r - 4)(r + 5) = 0 \] So, \( r = 4 \) or \( r = -5 \). 1. If \( r = 4 \): - First term: \( T_1 = 3 \) - Second term: \( T_2 = 3 \cdot 4 = 12 \) - Third term: \( T_3 = 3 \cdot 4^2 = 48 \) The first three terms are: \( 3, 12, 48 \) and the general term is: \[ T_n = 3 \cdot 4^{n-1} \] 2. If \( r = -5 \): - First term: \( T_1 = 3 \) - Second term: \( T_2 = 3 \cdot -5 = -15 \) - Third term: \( T_3 = 3 \cdot (-5)^2 = 75 \) The first three terms are: \( 3, -15, 75 \) and the general term is: \[ T_n = 3 \cdot (-5)^{n-1} \] So, depending on the common ratio chosen, you have either the terms \( 3, 12, 48 \) with \( r = 4 \), or \( 3, -15, 75 \) with \( r = -5 \).