Chapter 11.4, Problem 6CYU

To determine

To find the sum of infinite series(if it exists)

  $520+130+32.5+\mathrm{......}$

Answer to Problem 6CYU

  $\frac{2080}{3}$

Explanation of Solution

Given:

  $520+130+32.5+\mathrm{......}$

Concept Used:

 1. Sum of an infinite geometric series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $\left|r\right|<1$is:

  ${\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}=\frac{a}{1-r}$

 2. Sum of an infinite geometric series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $r>1$ is divergent(doesn’t exist).

Calculation:

In order to find the sum of infinite series

  $520+130+32.5+\mathrm{......}$

First find its first term and common ratio.

Note from the given geometric series note that first term is 520, and the common ratio ( r ) can be find by dividing second term by first term, as

  $r=\frac{a_2}{a_1}$

Since, here

  $a_2=130,a_1=520$

Thus,

  $\begin{array}{c}r=\frac{130}{520}\\ =\frac{1}{4}\\ =0.25\end{array}$

Since, $\left|r\right|<1$ .

Thus, sum of given infinite geometric progression can be find using the formula

  ${\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}=\frac{a}{1-r}$

As

  $\begin{array}{c}520+130+32.5+\mathrm{......}=\frac{520}{1-0.25}\\ =\frac{520}{0.75}\\ =\frac{52000}{75}\\ =\frac{2080}{3}\end{array}$

Thus,

  $520+130+32.5+\mathrm{......}=\frac{2080}{3}$