Question

A deposit of $1000 is made into an account at the beginning of each year for 30 years and earns 6% interest compounded annually. What is the value in the account at the end of thirtieth year?

I need this quick please!!!

Answer 1

I think it is about 13780 but I am not sure

Answer 2

Deposit made each year = $1000

Compound interest = 6%

Time period = 30 years

When a deposit is made each year with interest compounded annually, total amount after a certain time period is given as:

$P_N =\frac{d((1+\frac{r}{k})^{Nk}-1)}{\frac{r}{k}}$

where,

$P_N$ = total amount in account after N years.

d = deposit made

r = annual rate of interest in decimal form.

k = number times deposit made in one year.

$P_N =\frac{1000((1 +\frac{0.06}{1})^{(30*1)} -1)}{\frac{0.06}{1}}$

$P_N =\frac{1000((1+0.06)^{30}-1)}{0.06}$

$P_N =\frac{1000((1.06)^{30}-1)}{0.06}$

$P_N =\frac{1000(5.743 - 1)}{0.06}$

$P_N =\frac{1000 * 4.743}{0.06}$

$P_N =\frac{4743}{0.06}$

$P_N = 79050$

Hence, value in the account at the end of 30th year = $79050