Chapter 11.4, Problem 1BCYP

To determine

To find the sum of infinite geometric progression

  $100+50+25+\mathrm{......}$

Answer to Problem 1BCYP

200

Explanation of Solution

Given:

  $100+50+25+\mathrm{......}$

Concept Used:

1. Sum of an infinite geometric progression $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $0<r<1$is:

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

2. Sum of an infinite geometric progression $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $r>1$ is divergent.

Calculation:

In order to find the sum of infinite geometric progression

  $100+50+25+\mathrm{......}$

Use the formula

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

Note from the given geometric series note that first term is 100, and the common ratio ( r ) can be find by dividing second term by first term, as

  $r=\frac{a_2}{a_1}$

Since, here

  $a_2=50,a_1=100$

Thus,

  $\begin{array}{c}r=\frac{50}{100}\\ =0.5\end{array}$

Since, $0<r<1$ .

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

  $\begin{array}{c}100+50+25+\mathrm{......}=\frac{100}{1-0.5}\\ =\frac{100}{0.5}\\ =\frac{1000}{5}\\ =200\end{array}$

Thus,

  $100+50+25+\mathrm{......}=200$