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I'm having trouble understanding the solution for this problem

Susan invests Z at the end of each year for seven years at an effective annual interest rate of 5%. the interest credited at the end of each year is reinvested at an annual of 6%. The accumulated value at the end of 7 years is X.

Lori invests Z at the end of each year for fourteen years at an effective annual interest rate of 2.5%. the interest credited at the end of each year is reinvested at an annual of 3%. The accumulated value at the end of 14 years is Y.

Find $\frac YX$

The solution gives a formula:

$Accumulated Value = NZ + .05Z(Is)_{\overline6|.06}$

Where N is the number of years, and Z is the deposit amount.

I don't understand where this formula came from? Is there anyway to solve this question other than using this formula?

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user333888
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1 Answers1

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Observe the amount of money at each time \begin{align} t&=1 & &z\\ t&=2 & &z(1+i)+z=\color{blue}{2z}+\color{red}{iz}\\ t&=3 & &\color{blue}{2z}(1+i)+z=\color{blue}{3z}+\color{red}{2iz}\\ &\;\;\vdots\\ t&=n & &\color{blue}{(n-1)z}(1+i)+z=\color{blue}{nz}+\color{red}{(n-1)iz}\\ \end{align} The interest part will be subtracted each year and invested at rate $j$. So we have an increasing annuity $k(iz)$ for $k=1,\ldots,n-1$ for the interest part. Then the future value will be $$ FV=nz+iz\,(Is)_{\overline{n-1}|\,j} $$ So the future value for Susan will be $$ x=7z+0.05\,z\,(Is)_{\overline{6}|\,0.06} $$ and the future value for Lori will be $$ y=14z+0.025\,z\,(Is)_{\overline{13}|\,0.03} $$ Thus $$ \frac{x}{y}=\frac{7+0.05\,(Is)_{\overline{6}|\,0.06}}{14+0.025\,(Is)_{\overline{13}|\,0.03}} $$

alexjo
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