# Calculating the value of Annuities

Q: Instead of investing $3000$ at the end of $5$ years, and $\$4000$at the end of$10$years, Steve wishes to make regular monthly payments that will amount to the same total after$10$years. Determine the monthly payment if interest is compounded monthly at an annual rate of$4\%$Could anyone set up the annuity formula with the numbers from the question. Im not sure if what i did was correct: $$4000= R \frac{1- 1.003^{120}}{ 0.004/12}$$ The answer when you isloate for$R$should be$\$52.04$.

You have made some numerical errors and considered the contribution of $\$4000$only. Mathematically one can establish an equivalence between the investments (at the end of$5$years and at the end of$10$years) and a series of$120$monthly constant payments. Since there are$m=12$compounding periods per year, the (nominal) annual interest rate$r=4\%=0.04$indicates a monthly interest rate$i=\frac{r}{m}=\frac{4}{12}\%=\frac{0.04}{12}$. The hypothetical investment of$3000$at the end of$5$years ($60$months) will accumulate interest during$5$years ($60$months). Hence it's future value is $$F^{\prime }=3000\left( 1+i\right)^{60}=3000\left( 1+0.04/12\right) ^{60}\approx 3663.0.$$ Adding the second hypothetical investment$F^{\prime \prime }=4000$yields the total future value$F=F^{\prime }+F^{\prime \prime }=7663.0$at the end of$10$years. Let$A$(the annuity) denote each monthly payment. The payment at the end of month$k$increases to a future value of$F_{k}=A(1+i)^{n-k}$at the end of$n=120$months. Summing all these$F_{k}$the resulting geometric series of$n$payments, whose ratio is$c=1+i$, should be equal to$F\$, as a consequence of the equivalence mentioned above. Applying the formula for such a sum, we get \begin{equation*} F=\sum_{k=1}^{n}F_{k}=\sum_{k=1}^{n}A(1+i)^{n-k}=\sum_{j=1}^{n}Ac^{j-1}=A \frac{c^{n}-1}{c-1}=A \frac{(1+i)^{n}-1}{i}. \end{equation*} Numerically we obtain \begin{equation*} A=F\frac{i}{(1+i)^{n}-1}=7663.0\frac{\frac{0.04}{12}}{(1+\frac{0.04}{12} )^{120}-1}\approx 52.04, \end{equation*} which agrees with the answer you indicate.