# Sinking Fund (1 Viewer)

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#### gillgill

John brought a stamp for his collection. He agreed to pay a lump sum of $4000 after 5 years. Until then, he pays 6% simple interest semiannually on the$4000.
a) Find the amount of each semiannual interest payment.
b) John sets up a sinking fund so that enought money will be present to pay off the $4000. He will make annual payments into the fund. the accound pays 8% compounded annually. Find the amound of each payment. a) is it m=2 n=5x2=10 i=0.06/2=0.03 P=4000 and sub in A=P(1+i)^n and then solve for A interest=A-P therefore A-4000 so the interest is$1374.67???

b) i found that the amount of each payment is $681.83???? I am not really sure about my answers. #### HallsofIvy Science Advisor gillgill said: John brought a stamp for his collection. He agreed to pay a lump sum of$4000 after 5 years. Until then, he pays 6% simple interest semiannually on the $4000. a) Find the amount of each semiannual interest payment. b) John sets up a sinking fund so that enought money will be present to pay off the$4000. He will make annual payments into the fund. the accound pays 8% compounded annually. Find the amound of each payment.
a)
is it m=2
Is WHAT m= 2??
n=5x2=10
i=0.06/2=0.03
P=4000
and sub in A=P(1+i)^n
and then solve for A
interest=A-P
therefore A-4000
so the interest is $1374.67??? No, the problem specifically said "simple interest".$4000 at 6% interest would be (0.06)(4000)= $240 for a year and so$120 for 6 months. The problem asked you to find "the amount of each semiannual interest payment", not the total interest.

b) i found that the amount of each payment is $681.83???? I am not really sure about my answers. Assuming he makes the same payment each year, let P be that amount. The amount deposited at the beginning earns 8% compounded annually for 5 years. At the end of the 5 years, it will have increased to P(1.08)5. The amount he puts in at the beginning of the second year will earn 8% compounded annually for 4 years. At the end, it will have increased to P(1.08)4. The amount he puts in at the beginning of the third year will have increased to P(1.08)3. The amount he puts in at the beginning of the 4th year will have increased to P(1.08)2. The amount he puts in at the beginning of the 5th year will have increased to P(1.08). At the end of the 5th year, he will have P(1.08+ 1.082+ 1.083+ 1.084+ 1.085). That short enough to do on directly but you might recognize it as a geometric series with initial term a1 = P(1.08), common ratio r= 1.08, and summed from n= 0 to n= 4. The formula for such a sum is $a_1\frac{1- r^5}{1-r}$. With these values that will be [itex]P(1.08)\frac{1- 1.469}{1-1.08}= p(1.08)\frac{-0.469}{-0.08}= 6.34P[/tex]. Set that equal to$4000 and solve for P. I get \$631.32.

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