EXERCISE 4 The third term of a geometric sequence is 6 and the sixth term is \( \frac{3}{32} \). Determine the first three terms and the general term \( T_{n} \).

To solve for the first three terms of the geometric sequence, we can use the formula for the nth term: \( T_n = ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. From the information given: 1. The third term \( T_3 = ar^2 = 6 \) 2. The sixth term \( T_6 = ar^5 = \frac{3}{32} \) Now, we set up two equations based on these terms: 1. \( ar^2 = 6 \) 2. \( ar^5 = \frac{3}{32} \) We can express the second equation in terms of the first by dividing: \[ \frac{ar^5}{ar^2} = \frac{\frac{3}{32}}{6} \] This simplifies to: \[ r^3 = \frac{3}{32 \cdot 6} \] \[ r^3 = \frac{3}{192} = \frac{1}{64} \] Taking the cube root gives us: \[ r = \frac{1}{4} \] Now substituting \( r \) back to find \( a \): \[ ar^2 = 6 \] \[ a \left(\frac{1}{4}\right)^2 = 6 \] \[ a \cdot \frac{1}{16} = 6 \] \[ a = 6 \cdot 16 = 96 \] Thus, the first three terms are: - \( T_1 = a = 96 \) - \( T_2 = ar = 96 \cdot \frac{1}{4} = 24 \) - \( T_3 = ar^2 = 96 \cdot \left(\frac{1}{4}\right)^2 = 6 \) The general term \( T_n \) can now be expressed as: \[ T_n = 96 \left(\frac{1}{4}\right)^{n-1} \] So, the first three terms of the sequence are \( 96, 24, 6 \), and the general term is \( T_n = 96 \cdot \left(\frac{1}{4}\right)^{n-1} \).