Question

Mario deposits $25 at the end of each month for 4 years into an account that pays 9.6%/a compounded monthly. He then makes no further deposits and no withdrawals. Determine the balance 10 years after his last deposit.

Answer 1

This problem is related to annuity problem.

The formula for compound interest, including principle sum is:

$A=\frac{P\times[(1+\frac{r}{k})^{Nk}-1]}{\frac{r}{k}}$

where,

A = the balance in the account after N years.

P = the regular deposit (the amount you deposit each year, each month, etc.)

r = the annual interest rate in decimal form =$\frac{9.6}{100}=0.096$

k = the number of compounding period in one year

N = the number of years that interest is compounded

so, $A=\frac{25\times[(1+\frac{0.096}{12})^{4\times12}-1]} {\frac{0.096}{12}}$

$A=\frac{25\times[(1.008)^{48}-1]}{0.008}=\frac{25\times[1.46590404-1]}{0.008}=1,455.95012$

Thus, the amount after 4 years = $1,455.95012

but the there is no deposit and withdrawal after this and the amount 10 years after the last deposit is :

$A=P\times(1+\frac{R}{100})^T$

where,

A = total amount after the compounded interest on the principal amount

P = Principal amount

R = rate of interest compounded annually $=\frac{9.6}{12}$ %

T = time in years

$A=1455.95012\times(1+\frac{9.6}{12\times100})^{10\times12}$

$A=1455.95012\times(1.008)^{120}=1455.95012\times2.60173976$

A = $3,788.00332

Thus, the amount 10 years after the last deposit is : A = $3,788.00332