SOLUTION: Use graphical approximation techniques or an equation solver to approximate the desired interest rate. A person makes annual payments of $ 1000 into an ordinary annuity. At the end

Question 1120055: Use graphical approximation techniques or an equation solver to approximate the desired interest rate. A person makes annual payments of $ 1000 into an ordinary annuity. At the end of 5 years, the amount in the annuity is $ 5703.88. What annual nominal compounding rate has this annuity earned?
Found 2 solutions by Boreal, ankor@dixie-net.com:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
FV=Payment (1+i)^n)-1/i
5703.88=1000(1+i)^n-1/i
5.70388=(1+i)^5-1/i
Picking 0.03
(1+.03)^5=1.16, so 0.16/0.03=$5333.33
try -.04
1.2167-1/0.04=0.2167/0.04=5.416
Try 0.07
1.2762-1/.05=5.751, a little too big
try 0.069 and get 5.739
0.066 and get 5705.02
0.0659 will give the answer.
Interest rate is 6.59%
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A person makes annual payments of $ 1000 into an ordinary annuity.
At the end of 5 years, the amount in the annuity is $ 5703.88.
What annual nominal compounding rate has this annuity earned?
:
We can use the annuity formula
P() = Fv; where
P =payment
i = interest rate
t = periods
Fv = Final value
:
1000() = 5703.88
Put in the form that we can graph; i = x
1000() = 5703.88
divide both sides by 1000
=
Approx solution; x = .066 or 6.6% interest
= 5.704
multiply both sides by x
(1+x)^5 - 1 = 5.704x
A quadratic equation we can graph
(1+x)^5 - 5.704x - 1 = 0
:

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