SOLUTION: How much must be invested at the end of each year, for 4 years, to achieve an amount of $10000, if interest is earned at a rate of 6.25% per year, compounded annually?

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Question 1087785: How much must be invested at the end of each year, for 4 years, to achieve an amount of $10000, if interest is earned at a rate of 6.25% per year, compounded annually?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Let P = amount invested at the end of each year
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Year 1:
P dollars is invested at an interest rate of 6.25%, so r = 0.0625 in decimal form.
The compounding frequency is n = 1
The money is compounded for t = 3 years

So,
A = P*(1+r/n)^(n*t)
A = P*(1+0.0625/1)^(1*3)
A = P*1.199462890625
A = 1.199462890625*P

Let's call this A1, so A1 = 1.199462890625*P
-------------------------------
Year 2:
P dollars is invested at an interest rate of 6.25%, so r = 0.0625 in decimal form.
The compounding frequency is n = 1
The money is compounded for t = 2 years

So,
A = P*(1+r/n)^(n*t)
A = P*(1+0.0625/1)^(1*2)
A = P*1.12890625
A = 1.12890625*P

Let's call this A2, so A2 = 1.12890625*P
-------------------------------
Year 3:
P dollars is invested at an interest rate of 6.25%, so r = 0.0625 in decimal form.
The compounding frequency is n = 1
The money is compounded for t = 1 years

So,
A = P*(1+r/n)^(n*t)
A = P*(1+0.0625/1)^(1*1)
A = P*1.0625
A = 1.0625*P

Let's call this A3, so A3 = 1.0625*P
-------------------------------
Year 4:
P dollars is invested at an interest rate of 6.25%, so r = 0.0625 in decimal form.
The compounding frequency is n = 1
The money is compounded for t = 0 years

So,
A = P*(1+r/n)^(n*t)
A = P*(1+0.0625/1)^(1*0)
A = P*1
A = 1*P

Let's call this A4, so A4 = 1*P
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In summary, we have the following
A1 = 1.199462890625*P
A2 = 1.12890625*P
A3 = 1.0625*P
A4 = 1*P

Add up the values A1 through A4 to get
A1 + A2 + A3 + A4 = 1.199462890625*P + 1.12890625*P + 1.0625*P + 1*P
A1 + A2 + A3 + A4 = (1.199462890625 + 1.12890625 + 1.0625 + 1)*P
A1 + A2 + A3 + A4 = 4.390869140625*P
-------------------------------
We want the sum of the four results (A1 through A4) to add up to $10,000 as stated in the problem.
Set 4.390869140625*P equal to 10000 and solve for P
4.390869140625*P = 10000
4.390869140625*P/4.390869140625 = 10000/4.390869140625
P = 2277.45343341673
P = 2277.45

Rounded to the nearest penny, the amount you need to invest at the end of each year is $2,277.45

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A shortcut is to use the future value of an annuity formula
Plug in FV = 10000, i = 0.0625 and n = 4.
Solve for P

FV = P * [(1+i)^n - 1 ]/i
10000 = P * [(1+0.0625)^4 - 1 ]/0.0625
10000 = P * [(1.0625)^4 - 1 ]/0.0625
10000 = P * [1.27442932128906 - 1 ]/0.0625
10000 = P*(0.274429321289063)/0.0625
10000 = P*4.390869140625
10000 = 4.390869140625*P
4.390869140625*P = 10000
4.390869140625*P/4.390869140625 = 10000/4.390869140625
P = 2277.45343341674
P = 2277.45

Getting us the same answer as before.