Present and Accumulated Values of an Annuity-Immediate

Problem 15.1
Consider an investment of $5,000 at 6% convertible semiannually. How much can be withdrawn each half−year to use up the fund exactly at the end of 20 years?

Solution.
$$5,000 = Xa_{\overline{40|}0.03} \\ \implies X = 216.31$$

Problem 15.2
The annual payment on a house is $18,000. If payments are made for 40
years, how much is the house worth assuming annual interest rate of 6%?

Solution.
$$18,000a_{\overline{40|}{}} \\ = 270, 833.34$$

Problem 15.3
If $d = 0.05$, calculate $a_{\overline{12|}{}}$ .

Solution.
$$\text{Note: i = $\frac{d}{1-d}$} \\ \text{Substituting the value of d, we have} \\ i = 5.26\% \text{Solving for $a_{\overline{12|}{}}$, we have} \\ a_{\overline{12|}{}} = 8.733$$

Problem 15.4
Calculate the present value of 300 paid at the end of each year for 20 years
using an annual effective interest rate of 8%.

Solution.
$$300a_{\overline{20|}{}} = 2,945.44$$

Problem 15.5
If $a_{\overline{n|}{}} = x$ and $a_{\overline{2n|}{}} = y$, express $d$ as a function of $x$ and $y$.

Solution.
$$a_{\overline{2n|}{}} = a_{\overline{n|}{}} + v^na_{\overline{n|}{}} \\ y = x + v^nx \\ \frac{y-x}{x} = v^n \\ 1 - \frac{y-x}{x} = 1 - v^n \\ \frac{x-(y-x)}{x} = 1 - v^n \\ \frac{2x-y}{x} = 1 - v^n \\ \frac{2x-y}{ix} = \frac{1 - v^n}{i} \\ \frac{2x-y}{ix} = a_{\overline{n|}{}} \\ \frac{2x-y}{ix} = x \\ \frac{2x-y}{\frac{d}{1-d}x} = x \\ 2x-y = x^2\frac{d}{1-d} \\ (1-d)(2x-y) = x^2d \\ (2x-y)-d(2x-y) = x^2d \\ 2x-y = x^2d+d(2x-y) \\ 2x-y = d(x^2 + 2x-y) \\ d = \frac{2x-y}{x^2 + 2x-y}$$

Problem 15.6
(a) Given: $a_{\overline{7|}{}} = 5.153$, $a_{\overline{11|}{}} = 7.036$, $a_{\overline{18|}{}} = 9.180$. Find $i$.
(b) You are given that $a_{\overline{n|}{}} = 10.00$ and $a_{\overline{3n|}{}} = 24.40$. Determine $a_{\overline{4n|}{}}$ .

Solution.
$$\text{a.) Recall: } a_{\overline{18|}{}} = a_{\overline{7|}{}} + v^7a_{\overline{11|}{}} \\ \text{Substituting the given values, we have} \\ 9.180 = 5.153 + (1+i)^{-7}(7.036) \\ \implies i = 8.29\%$$