Math Problem Statement

Suppose an annuity will pay $16,000 at the beginning of each year for the next 8 years. How much money is needed to start this annuity if it earns 6.3%, compounded annually? (Round your answer to the nearest cent.)

Solution

To determine the present value of an annuity where payments are made at the beginning of each year, we need to use the formula for the present value of an annuity due. The formula for an annuity due (which pays at the start of each period) is:

PV=P×[1(1+r)nr]×(1+r)PV = P \times \left[ \frac{1 - (1 + r)^{-n}}{r} \right] \times (1 + r)

Where:

  • PVPV is the present value of the annuity.
  • PP is the payment amount ($16,000).
  • rr is the interest rate per period (6.3% = 0.063).
  • nn is the number of periods (8 years).

Let’s plug in the values and calculate the present value.

PV=16,000×[1(1+0.063)80.063]×(1+0.063)PV = 16,000 \times \left[ \frac{1 - (1 + 0.063)^{-8}}{0.063} \right] \times (1 + 0.063)

I'll calculate this now.The amount of money needed to start the annuity is $104,373.50 (rounded to the nearest cent).

Would you like further details or clarification on any part of this calculation?

Here are 5 related questions to consider:

  1. How would the present value change if the interest rate were lower, say 5%?
  2. What is the difference between an annuity due and an ordinary annuity?
  3. How much would the annuity pay in total over 8 years?
  4. How would the calculation change if payments were made at the end of each year instead?
  5. What if the annuity paid quarterly instead of annually?

Tip: An annuity due always yields a higher present value than an ordinary annuity, because payments are made earlier (at the start of each period).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Annuities
Present Value
Interest Rate
Compounding

Formulas

Present value of an annuity due: PV = P * [(1 - (1 + r)^-n) / r] * (1 + r)

Theorems

Annuity Due Theorem

Suitable Grade Level

College Level / Advanced High School (Grades 11-12)