Question

What is the sum of the first eight terms of a geometric series whose first term is 3 and whose common ratio is .5 ?

Answer 1

$\bf \qquad \textit{sum of a finite geometric sequence} \\ \quad \\ S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term}\\ a_1=\textit{first term}\\ r=\textit{common ratio} \end{cases} \\ \quad \\ S_8=3\left( \cfrac{1-0.5^8}{1-0.5} \right)$

Answer 2

Answer:  The required sum of first eight terms of the given geometric series is 5.98.

Step-by-step explanation:  We are given to find the sum of first eight terms of a geometric series whose first term is 3 and whose common ratio is 0.5.

We know that

the sum of first n terms of a geometric series with first term a and common ratio r is given by

$S_n=\dfrac{a(1-r^n)}{1-}.$

For the given geometric series, we have

first term, a = 3  and  common ratio, r = 0.5.

So, the sum of first eight terms of the given geometric series will be

$S_8\\\\\\=\dfrac{a(1-r^8)}{1-r}\\\\\\=\dfrac{3\left(1-\left(\frac{1}{2}\right)^8\right)}{1-\frac{1}{2}}\\\\\\=\dfrac{3\left(1-\frac{1}{256}\right)}{\frac{1}{2}}\\\\\\=3\times2\times\dfrac{256-1}{256}\\\\\\=3\times\dfrac{255}{128}\\\\\\=\dfrac{765}{128}\\\\=5.98.$

Thus, the required sum of first eight terms of the given geometric series is 5.98.