Find the value of the ordinary annuity at the end of the indicated time period. The payment R, frequency of deposits m (which is the same as the frequency of compounding), annual interest rate r, and time period t are given.
amount 1,000; quarterly; 6.2%, five years
How do I solve this?
First I took r/m=.062/4=.0155
Then I took T=5 years and times it by 4 (compound quarterly) =20
Then I entered
1000(1+.0155/4)^20-1/(.0155)
What do I do from here?
Thanks!
1000(1+.0155/4)^20-1/(.0155)
careful with brackets, secondly you already did .062/4 , so why are you dividing .0155 by 4 ??
amount = 1000 ( 1.0155^20 - 1)/.0155
= 1000( 23.237857...)
= $ 23237.86
To solve for the value of the ordinary annuity, you've correctly calculated the value for r/m as 0.0155, and T as 20 (since the time period is 5 years compounded quarterly). However, you made a small mistake in the equation.
The correct formula to find the value of the ordinary annuity is:
P = R * [(1 + r/m)^(mt) - 1] / (r/m)
Where:
P is the present value (amount),
R is the payment (annuity),
r is the annual interest rate as a decimal,
m is the frequency of deposits (compounding), and
t is the time period in years.
Replacing the values into the formula:
P = 1000 * [(1 + 0.0155/4)^(4*5) - 1] / (0.0155/4)
Now we can simplify the equation:
P = 1000 * [(1.003875)^20 - 1] / 0.003875
You can use a calculator to evaluate the expression inside the brackets (1.003875^20) which is approximately equal to 1.080618315.
Now the equation becomes:
P = 1000 * (1.080618315 - 1) / 0.003875
P = 1000 * 0.080618315 / 0.003875
P = 20690.06
Therefore, the value of the ordinary annuity at the end of the indicated time period is approximately $20,690.06.