15. The \( n^{\text {th }} \) term of a geometric series is \( \mathrm{T}_{n}=x(x+1)^{n-1} \). (a) Determine the constant ratio in terms of \( x \) and in simplest form. (b) If \( x=1 \), determine the sum of the first 25 terms of the series. 16. The sequence 12; \( x ; \ldots \) is a convergent geometrie sequence in which the sum to infi is equal to 24 . (a) Determine the value of \( x \). (b) Hence, determine which term of the sequence is equal to \( \frac{3}{128} \).

The constant ratio of the geometric series can be found by comparing consecutive terms. The first term is \( \mathrm{T}_1 = x \) and the second term is \( \mathrm{T}_2 = x(x + 1) \). Thus, the common ratio \( r \) is: \[ r = \frac{\mathrm{T}_2}{\mathrm{T}_1} = \frac{x(x + 1)}{x} = x + 1 \] For the sum of the first 25 terms when \( x = 1 \), we substitute into the formula: \[ \mathrm{T}_n = 1(1 + 1)^{n - 1} = 2^{n - 1} \] The sum \( S_n \) of the first \( n \) terms of a geometric series is given by: \[ S_n = a \frac{r^n - 1}{r - 1} \] Using \( a = 1 \) and \( r = 2 \): \[ S_{25} = 1 \cdot \frac{2^{25} - 1}{2 - 1} = 2^{25} - 1 \] Now onto the convergent geometric sequence! The sum to infinity of a geometric sequence is calculated using: \[ S_\infty = \frac{a}{1 - r} \] Given that \( S_\infty = 24 \) and \( a = 12 \), we set it up: \[ 24 = \frac{12}{1 - r} \] This leads to \( 1 - r = \frac{12}{24} = \frac{1}{2} \) and thus \( r = \frac{1}{2} \). Now, using \( a \) and the common ratio \( r \), the second term \( x \) can be found: \[ x = 12 \cdot r = 12 \cdot \frac{1}{2} = 6 \] To find the term that equals \( \frac{3}{128} \), we have: \[ \mathrm{T}_n = 12 \cdot \left(\frac{1}{2}\right)^{n - 1} \] Setting the term equal to \( \frac{3}{128} \): \[ 12 \cdot \left(\frac{1}{2}\right)^{n - 1} = \frac{3}{128} \implies \left(\frac{1}{2}\right)^{n - 1} = \frac{3}{12 \cdot 128} = \frac{1}{512} \implies \frac{1}{2^{n - 1}} = \frac{1}{2^9} \] Thus, \( n - 1 = 9 \) or \( n = 10 \). So the 10th term equals \( \frac{3}{128} \).