Question

A rich man promises to give you $1000 on September 1, 2011. Each day thereafter he will give you $\frac{9}{10}$of what he gave you the previous day. What is the

first date on which the amount you receive is less than 1¢? How much will you have received when this happens?

Short answer

We will receive $9999.92

Step-by-step solution

Given informtion

We are given $a=1000,r=0.9$

We find the value of n

For $a_n$exactly to be 0.01 we would have

role="math" localid="1647865193081" $$\begin{gathered} 0.01=a_1{(0.09)}^{n-1} \\ 0.01=1000{(0.09)}^{n-1} \\ 0.00001=0.{09}^{n-1} \\ Thereforen=110.27 \end{gathered}$$

Calculate S

We use n=111

$$\begin{gathered} S=a_1\frac{1-r^{111}}{1-r} \\ S=\left(1000\right)\frac{1-0.9^{111}}{1-0.9} \\ S=9999.92 \end{gathered}$$

Conclusion

We will receive $9999.92