For all values of \(n\) the sum of \(n\) terms of a geometrical progression is \(\dfrac{3^n - 2^n}{2^{n-5}}\). Find its first term and common ratio.
A geometric progression takes the form \(a, ar, ar^2, \dotsc\), where \(a\) is the first term and \(r\) is the common ratio. We are told that, for each \(n \ge 1\), \[ a + ar + \dotsb + ar^{n-1} = \frac{3^n - 2^n}{2^{n-5}}. \]
When \(n = 1\), this becomes \[ a = \frac{3^1 - 2^1}{2^{1-5}} = \frac{3 - 2}{2^{-4}} = 1 \times 2^4 = 16. \]
When \(n = 2\), this becomes
\[a + ar = \frac{3^2 - 2^2}{2^{2-5}} \implies 16 + 16r = \frac{9 - 4}{2^{-3}} = 5 \times 2^3 = 40. \]
Thus the second term \(ar = 40 - 16 = 24\), which means that \(r = \dfrac{24}{16} = \dfrac{3}{2}\).
