Question

Regular annual deposit are made into a savings account at the end of each year for 16 years the value of the first deposit is R8000 and thereafter deposit are increased each year from the second deposit onwards at a rate of 3%p.a if the interest rate earned on the savings account is 4.4% p.a compounded monthly what is the future value of the growing annuity?

Answer 1

If the time starts at $t_0=0$, first deposit will be made at the end of the first year, i.e at $t_1=1$ year, second deposit - at $t_2=2$ year, and so on until 16th deposit will be made at the end of 16th year at $t_{16}=16$ years.

Deposit made at $t_1=1$ year is equial to R 8000, deposit made at $t_2=2$ will be 3% bigger i.e. $8000*1.03$ and so on, deposit made at $t_{16}=16$ will be equial to $8000*1.03^{15}$ .

We need to find future value of all deposits, i.e. the value that those deposits will have at $t_{16}=16$ given interest rate on savings account and the fact that compounding is monthly.

Future value of first deposit will be:

$8000*(1+\frac{0.044}{12})^{(16-1)*12}=8000*(1+\frac{0.044}{12})^{15*12}$

Future value of the second deposit will be: $8000*1.03*(1+\frac{0.044}{12})^{(16-2)*12}=8000*1.03*(1+\frac{0.044}{12})^{14*12}$

Future value of 15th deposit will be:

$8000*1.03^{14}*(1+\frac{0.044}{12})^{(16-15)*12}=8000*1.03^{14}*(1+\frac{0.044}{12})^{12}$

Future value of the 16th deposit will be equial to its present value at $t_{16}=16$

$8000*1.03^{15}$

Therefore future value of all deposits made (=growing annuity) will be the sum of mentioned above components:

$8000*(1+\frac{0.044}{12})^{15*12}+8000*1.03*(1+\frac{0.044}{12})^{14*12}+ ....+\\+8000*1.03^{14}*(1+\frac{0.044}{12})^{12}+8000*1.03^{15}=\\=8000*[(1+\frac{0.044}{12})^{15*12}+1.03*(1+\frac{0.044}{12})^{14*12}+...+1.03^{14}*(1+\frac{0.044}{12})^{12}+1.03^{15}]$

In brackets we can see progression with following properties:

first item $b_1=1.03^{15}$ , ratio $q=\frac{(1+\frac{0.044}{12})^{12}}{1.03}$ , number of elements $n=16$.

Its sum can be calculated by formula:$b_1*\frac{q^n-1}{q-1}=1.03^{15}*\frac{\frac{(1+\frac{0.044}{12})^{12*16}}{{1.03^{16}}}-1}{\frac{(1+\frac{0.044}{12})^{12}}{{1.03}}-1}= \frac{(1+\frac{0.044}{12})^{12*16}-1.03^{16}}{(1+\frac{0.044}{12})^{12}-1.03}\approx27.8231$

Which means that future value of growing annuity will be equial to $8000*27.8231=222584.8$