Chapter 11, Problem 11CR

To determine

To select: The correct option which shows sum of infinite geometric series with $64+48+36+27+\cdot \cdot \cdot$

(a). $\frac{256}{7}$

(b). $\frac{256}{3}$

(c). $256$

(d). no sum

Answer to Problem 11CR

The correct option is (c).

The sum of infinite geometric series with $64+48+36+27+\cdot \cdot \cdot$ is $256$ .

Explanation of Solution

Given information:

  $64+48+36+27+\cdot \cdot \cdot$

Formula used:

The sum of infinite geometric series: $S=\frac{a\left(Sub1\right)}{\left(1-r\right)\left(\left|r\right|<1\right)}$

Calculation:

The sum of the infinite geometric series

  $\begin{array}{l}64\times \frac{3}{4}=48\\ 48\times \frac{3}{4}=36\\ 36\times \frac{3}{4}=27\\ 64\times {\left(\frac{3}{4}\right)}^0=64\times 1=64\\ 64\times {\left(\frac{3}{4}\right)}^1=64\times \frac{3}{4}=48\\ 64\times {\left(\frac{3}{4}\right)}^2=64\times \frac{9}{16}=36\\ 64\times {\left(\frac{3}{4}\right)}^3=64\times \frac{27}{64}=27\\ 64\times {\left(\frac{3}{4}\right)}^2=64\times \frac{81}{256}=20.25\\ \end{array}$

The formula of sum of infinite geometric series:

  $S=\frac{a\left(Sub1\right)}{\left(1-r\right)\left(\left|r\right|<1\right)}$

Put the value of $r=3/4$ and $a\left(sub1\right)=64$ :

  $\begin{array}{l}S=\frac{64}{\left(1-\frac{3}{4}\right)}\\ S=\frac{64}{\left(1/4\right)}\\ S=256\end{array}$

The correct option is (c).

The sum of infinite geometric series with $64+48+36+27+\cdot \cdot \cdot$ is $256$ .