Question

Find the sum of the infinite geometric series.

$48+24+12+6++3+\frac{3}{2}+...$

Short answer

Sum is$96$.

Step-by-step solution

Step 1. Given information

Infinite geometric series is $48+24+12+6++3+\frac{3}{2}+...$

Step 2. 

Consider the infinite geometric series: $48+24+12+6+3+\frac{3}{2}+...$

Let $a,r$denote the first term and the common ratio respectively.

$\begin{array}{rcl}r& =& \frac{24}{48}\\ & =& \frac{1}{2}\end{array}$

$S_n=\frac{a}{1-r}$where $S_n$denotes the sum.

$\begin{array}{rcl}{S}_n& =& \frac{48}{1-{\displaystyle \frac{1}{2}}}\\ & =& 96\end{array}$