Calculate Interest & Payments - John's $4000 Lump Sum Investment

[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]

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 a_{1= 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].<br /> <br /> Set that equal to $4000 and solve for P. I get $631.32.[/itex]}}