Question

Question #ece03

Answer 1

Considering a GP of the type:
`a,ar,ar^2,ar^3,...ar^n`
where:
`a` is the first term;
`r` is the common ratio.
We have:
`ar^3=56` and
`ar^5=7/8`
From the first:
`a=56/r^3`
Substituting in the second you get:
`(56r^5)/r^3=7/8` and
`r=1/8`
`a=28672`

and

`sum_(k=0)^ooar^k=a/(1-r)=28672/(1-1/8)=32768`