ESTION 3 Given that in a geometric sequence \( \mathrm{T}_{9}=768 \) and \( \mathrm{T}_{13}=12288 \). Determine the value(s) of the common ratio and the first term of the sequence. The sum to infinity of a convergent geometric series is \( \frac{54}{19} \). The sum to infinity of the same series calculated from the \( 3^{\text {rd }} \) term is \( \frac{24}{19} \). \( 3.2 .1 \quad \) Calculate the sum of the first two terms of the series. \( 3.2 .2 \quad \) Show that: \( a=\frac{30}{19(1+r)} \) \( 3.2 .3 \quad \) Determine the value of \( r \), if \( r>0 \)

To tackle the equations \( T_9 = ar^8 = 768 \) and \( T_{13} = ar^{12} = 12288 \), divide the second by the first: \( \frac{ar^{12}}{ar^8} = \frac{12288}{768} \). This reduces to \( r^4 = 16 \), giving \( r = 2 \) (since \( r > 0 \)). Plugging \( r \) back, solve for \( a \) from \( 768 = ar^8 \) to find \( a = 3 \). With the sum to infinity \( S_\infty = \frac{a}{1 - r} \) calculated as \( \frac{54}{19} \) for the series and \( S_\infty^{(3)} = \frac{ar^2}{1 - r} = \frac{24}{19} \) from the third term, equate to deduce an equation that establishes a relationship between \( a \) and \( r \). This can lead to finding the values of \( a \) and \( r \) by setting up the corresponding equations. By summing up the first two terms \( S_2 = a + ar = a(1 + r) \) and using information from the total series, we can compute the sum succinctly. When \( a = \frac{30}{19(1+r)} \) emerges, substitute known values of \( a \) and use \( r \) to solve for it clearly! Now, bring it home—substituting \( a \) and \( r \) into your equations can yield the proper value of \( r \). Let’s apply that mathematical magic!