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

In a geometric sequence, the terms can be described as \( T_n = ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. Given that \( T_3 = ar^2 = 6 \) and \( T_6 = ar^5 = \frac{3}{32} \), we can set up the equations to determine \( a \) and \( r \). Dividing the equations, we find \( \frac{T_6}{T_3} = \frac{ar^5}{ar^2} = r^3 = \frac{3/32}{6} \). Solving for \( r \), we discover \( r = -\frac{1}{4} \). Plugging \( r \) back into one of the equations, we can find \( a = 12 \). Hence, the first three terms are \( 12, -3, 6 \), and the general term can be expressed as \( T_n = 12 \left( -\frac{1}{4} \right)^{n-1} \). To double-check our work, let's verify some values: for \( n = 3 \), \( T_3 = 12 \left( -\frac{1}{4} \right)^2 = 6 \) (correct!), and for \( n = 6 \), \( T_6 = 12 \left( -\frac{1}{4} \right)^5 = -\frac{3}{32} \) (also correct!). All checks are in alignment, so we can confidently conclude the sequence parameters!