QUESTION 2 2.1 In a geometric sequence, \( T_{6}=-243 \) and \( T_{3}=72 \). Determine: 2.1.1 The constant ratio. 2.1.2 The first term. 2.1.3 The sum of the first 10 terms 2.2 Consider the series \( 16 k+8 k^{2}+4 k^{3}+\cdots \) 2.2.1 For which value(s) of \( k \) will the series converge? 2.2.2 Calculate the sum to infinity if \( k=-1,5 \).

The formula for the n-th term of a geometric sequence is \( T_n = a \cdot r^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio. Plugging in the values you provided, you can find \( r \) by using the equations derived from \( T_6 = a \cdot r^{5} = -243 \) and \( T_3 = a \cdot r^{2} = 72 \). This gives you two equations to solve for \( r \) and \( a \). For the series \( 16k + 8k^2 + 4k^3 + \cdots \), you can identify it as a geometric series with the first term \( a = 16k \) and common ratio \( r = \frac{8k^2}{16k} = \frac{1}{2}k \). The series converges when \( |r| < 1 \), which translates to \( | \frac{1}{2}k | < 1 \) or \( -2 < k < 2 \). The sum to infinity can be calculated using the formula \( S = \frac{a}{1 - r} \) for \( k = -1.5 \), leading to the final answer through these substitutions.