Chapter 8.3, Problem 80E

To determine

The sum of the given infinite geometric series.

Answer to Problem 80E

  $8+6+\frac{9}{2}+\frac{27}{8}+\mathrm{...}=32$

Explanation of Solution

Given information: The infinite geometric series $8+6+\frac{9}{2}+\frac{27}{8}+\mathrm{...}$

Concept used:

In a sequence $a_1,a_2,a_3,a_4,\dots \dots ,a_n,\dots$ if ratio of the consecutive terms is the same then it is defined as Geometric Sequence.

So, a number r is defined as the common ratio of the geometric sequence if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\dots \dots =r.$

The sum of an infinite geometric series $a_1+a_{1}r+a_1{r}^2+a_1{r}^3+\cdots +a_1{r}^{n-1}+\cdots$ is $S={\displaystyle {\sum }_{i=0}^{\infty }{a}_1{r}^i=\frac{a_1}{1-r}}$ where $\left|r\right|<1$ .

Calculation:

Consider the series $8+6+\frac{9}{2}+\frac{27}{8}+\mathrm{...}$

Here the first term $t_1=a=8$ , second term $t_2=6$

Then the common ratio $r=\frac{t_2}{t_1}=\frac{6}{8}=\frac{3}{4}$

Now,

  $\begin{array}{c}8+6+\frac{9}{2}+\frac{27}{8}+\mathrm{...}=\frac{a}{1-r}\\ =\frac{8}{1-\frac{3}{4}}\\ =32\end{array}$

Therefore, $8+6+\frac{9}{2}+\frac{27}{8}+\mathrm{...}=32$

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