Question

What is the sum of a 6-term geometric series if the first term is 23 and the last term is 1

Answer 1

Let the first term of GP be a and common ratio be r. Given a = 32 & T6 = 1.
$i.e. \: a {r}^{6 - 1} = 1 \\ 32 \times {r}^{5} = 1. \: \: \: \: \: {r}^{5} = \frac{1}{32} \\ r = \frac{1}{2}$
Sum of n terms of GP is
$\frac{a(1 - {r}^{n}) }{1 - r} \: \: \: for \: |r| < 1$
Sum of 6 terms is
$\frac{32 \times (1 - {( \frac{1}{2} )}^{6} )}{1 - \frac{1}{2} } = \frac{32 \times \frac{63}{64} }{ \frac{1}{2} } = 63$