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 find the first three terms of the geometric sequence, we start with the information given: the third term \( T_3 = 6 \) and the sixth term \( T_6 = \frac{3}{32} \). In a geometric sequence, each term is found by multiplying the previous term by a common ratio \( r \). We can express these terms as: - \( T_3 = ar^2 = 6 \) - \( T_6 = ar^5 = \frac{3}{32} \) Where \( a \) is the first term of the sequence. We can create a system of equations from these expressions: 1. \( ar^2 = 6 \) (Equation 1) 2. \( ar^5 = \frac{3}{32} \) (Equation 2) Now divide Equation 2 by Equation 1: \[ \frac{ar^5}{ar^2} = \frac{\frac{3}{32}}{6} \] This simplifies to: \[ r^3 = \frac{3}{32 \cdot 6} = \frac{3}{192} = \frac{1}{64} \] Taking the cube root of both sides gives: \[ r = \frac{1}{4} \] Now that we know \( r \), we can substitute the value of \( r \) back into Equation 1 to solve for \( a \): \[ ar^2 = 6 \implies a \left(\frac{1}{4}\right)^2 = 6 \implies a \cdot \frac{1}{16} = 6 \implies a = 6 \cdot 16 = 96 \] Now we can find the first three terms: 1. \( T_1 = a = 96 \) 2. \( T_2 = ar = 96 \cdot \frac{1}{4} = 24 \) 3. \( T_3 = ar^2 = 96 \cdot \left(\frac{1}{4}\right)^2 = 6 \) Thus, the first three terms are \( 96, 24, 6 \). To find the general term \( T_n \): \[ T_n = ar^{n-1} = 96 \left(\frac{1}{4}\right)^{n-1} \] This simplifies to: \[ T_n = 96 \cdot \frac{1}{4^{n-1}} = \frac{96}{4^{n-1}} \] So, the general term is \( T_n = \frac{96}{4^{n-1}} \).