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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!!!

2 Answers

4 votes
I think it is about 13780 but I am not sure
User Andrewrk
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2 votes

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 =(d((1+(r)/(k))^(Nk)-1))/((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 =(1000((1 +(0.06)/(1))^((30*1)) -1))/((0.06)/(1))


P_N =(1000((1+0.06)^(30)-1))/(0.06)


P_N =(1000((1.06)^(30)-1))/(0.06)


P_N =(1000(5.743 - 1))/(0.06)


P_N =(1000 * 4.743)/(0.06)


P_N =(4743)/(0.06)


P_N = 79050

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

User GeoBeez
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