Review question

# If a GP has $S_n=(3^n - 2^n)/2^{n-5}$, what's its common ratio? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6608

## Solution

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 $r = \dfrac{24}{16} = \dfrac{3}{2}$.