Question

Which of the following choices, A or B, results in more money?

A: To receive \(1000 on day 1, \)999 on day 2, \(998 on day 3, with the process to end after 1000 days

B: To receive \)1 on day 1, \(2 on day 2, \)4 on day 3, for

19 days

Short answer

Choice B results in more money

Step-by-step solution

Given information

We are given two choices

A: To receive $1000 on day 1, $999 on day 2, $998 on day 3, with the process to end after 1000 days

B: To receive $1 on day 1, $2 on day 2, $4 on day 3, for

19 days

We use arithmetic progression to find the amount of money

Here the first term is 1000 and common difference is 1.

Hence

$S_{1000}=\frac{1000}{2}[a+a_{1000}]$

As the common difference is 1 and first term is 1000 hence the 1000th term is 1

Therefore

$$\begin{gathered} S_{1000}=500(1000+1) \\ {S}_{1000}=500500 \end{gathered}$$

Hence choice A will give you $500500

Calculate choice B

in case B we have, a=1 and r-2

Also we have to find the sum of first 19 terms

hence we get,

$$\begin{gathered} S_{19}=a_1\times \frac{1-r^{19}}{1-r} \\ {S}_{19}=1\times \frac{1-2^{19}}{1-2} \\ {S}_{19}=524287 \end{gathered}$$

Hence choice B gives you $524827

Conclusion

Choice B results in more money