Chapter 11.3, Problem 42PPS

To determine

Find the sum of the geometric series with the information: $a_1=36;r=\frac{1}{3},n=8$

Answer to Problem 42PPS

Sum of the geometric series is ${\text{S}}_{\text{n}}=53\frac{241}{243}$

Explanation of Solution

Given:

Find the sum of the geometric series with the information: $a_1=36;r=\frac{1}{3},n=8$

Concept Used:

The general formula for a geometric series is ${\text{S}}_{\text{n}}=\frac{a_1(1-r^n)}{1-r}$

Calculation:

Given: $a_1=36;r=\frac{1}{3},n=8$

  $\begin{array}{l}{a}_1=36;r=\frac{1}{3},n=8;\text{Find}{\text{S}}_{\text{n}}\\ {\text{S}}_{\text{n}}=\frac{a_1(1-r^n)}{1-r}\\ {\text{S}}_{\text{n}}=\frac{36[1-{(\frac{1}{3})}^8]}{1-\frac{1}{3}}=\frac{36·(1-\frac{1}{6561})}{\frac{2}{3}}=\frac{36·\frac{6560}{6561}}{\frac{2}{3}}=36·\frac{6560}{6561}·\frac{3}{2}=\frac{13120}{243}=53\frac{241}{243}\\ {\text{S}}_{\text{n}}=53\frac{241}{243}\end{array}$

Thus, sum of the geometric series is ${\text{S}}_{\text{n}}=53\frac{241}{243}$