Chapter 9.3, Problem 78E

To determine

The sum of the infinite geometric series $-\frac{125}{36}+\frac{25}{6}-5+6-\mathrm{.............}$ .

Answer to Problem 78E

The sum of given infinite geometric series is $S=-\frac{625}{396}$

Explanation of Solution

Given information :

Given series is $-\frac{125}{36}+\frac{25}{6}-5+6-\mathrm{.............}$

As we have $-\frac{125}{36}+\frac{25}{6}-5+6-\mathrm{.............}$

  $\begin{array}{l}=-\frac{125}{36}+\frac{25}{6}-5+6-\mathrm{.............}\\ =-\frac{125}{36}\left(1-\frac{6}{5}+\frac{36}{25}-\frac{6\times 36}{125}-\mathrm{.............}\right)\\ =-\frac{125}{36}{{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{6}{5}\right)}}^{n-1}\end{array}$

Thus, the given series is a infinite geometric series with common ratio $r=-\frac{6}{5},a_1=1.$ ,

Therefore, the sum of infinite series is given by $S=\left(-\frac{125}{36}\right)\left(\frac{a_1}{1-r}\right)$

  $\begin{array}{l}\Rightarrow S=\left(-\frac{125}{36}\right)\left(\frac{1}{1-\left(-\frac{6}{5}\right)}\right)\\ =\left(-\frac{125}{36}\right)\left(\frac{1}{\raisebox{1ex}{$\left(5+6\right)$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right)\\ =\left(-\frac{125}{36}\right)\frac{5}{11}\\ \therefore S=-\frac{625}{396}\end{array}$

Hence, the sum of given infinite geometric series is $S=-\frac{625}{396}$