1. **Problem Statement:** We are given several annuity problems with different parameters and periodic payments. The goal is to verify or correct the periodic payment calculations. 2. **Formulas and Concepts:** - For an annuity, the periodic payment $P$ can be calculated using the formula for the present value of an annuity: $$A_{PV} = P \times \frac{1 - (1 + i)^{-n}}{i}$$ where $i$ is the interest rate per compounding period and $n$ is the total number of compounding periods. - For future value annuities, the formula is: $$A = P \times \frac{(1 + i)^n - 1}{i}$$ - Important: Interest rate per period $i$ = annual interest rate divided by number of compounding periods per year. - Total periods $n$ = number of years multiplied by number of compounding periods per year. 3. **Verification and Calculation:** **Simple (FV) Annuity:** - Amount $A = 7000$ - Interest rate = 8% annually, compounded semi-annually, so $i = 0.08/2 = 0.04$ - Term = 6 years, so $n = 6 \times 2 = 12$ - Given periodic payment $P = 745.87$ Check if $A = P \times \frac{(1 + i)^n - 1}{i}$: $$745.87 \times \frac{(1 + 0.04)^{12} - 1}{0.04} = 745.87 \times \frac{1.601032 - 1}{0.04} = 745.87 \times 15.0258 = 11200.5$$ This does not equal 7000, so the payment seems incorrect. Calculate correct $P$: $$P = \frac{A \times i}{(1 + i)^n - 1} = \frac{7000 \times 0.04}{1.601032 - 1} = \frac{280}{0.601032} = 465.9$$ **Simple (PV) Annuity:** - Amount $A_{PV} = 12000$ - Interest rate = 8.8% annually, compounded quarterly, so $i = 0.088/4 = 0.022$ - Term = 5 years, so $n = 5 \times 4 = 20$ - Given periodic payment $P = 736.90$ Check if $A_{PV} = P \times \frac{1 - (1 + i)^{-n}}{i}$: $$736.90 \times \frac{1 - (1 + 0.022)^{-20}}{0.022} = 736.90 \times \frac{1 - 0.647}{0.022} = 736.90 \times 15.14 = 11156.5$$ This is less than 12000, so payment is low. Calculate correct $P$: $$P = \frac{A_{PV} \times i}{1 - (1 + i)^{-n}} = \frac{12000 \times 0.022}{1 - 0.647} = \frac{264}{0.353} = 748.58$$ **General (A) Annuity:** - Amount $A = 120000$ - Interest rate = 6% annually, compounded quarterly, so $i = 0.06/4 = 0.015$ - Term = 10 years, so $n = 10 \times 4 = 40$ - Payment interval monthly means payments are monthly but compounding quarterly; adjust accordingly. Since payments are monthly but compounding quarterly, effective interest per month is: $$i_{month} = (1 + 0.015)^{1/3} - 1 = 0.004978$$ Total months $n = 10 \times 12 = 120$ Calculate $P$: $$P = \frac{A \times i_{month}}{(1 + i_{month})^n - 1} = \frac{120000 \times 0.004978}{(1.004978)^{120} - 1}$$ Calculate denominator: $$(1.004978)^{120} = e^{120 \times \ln(1.004978)} = e^{0.595} = 1.813$$ So denominator = $1.813 - 1 = 0.813$ Calculate $P$: $$P = \frac{120000 \times 0.004978}{0.813} = \frac{597.36}{0.813} = 734.9$$ Given payment is 402.24, which is too low. **General (A_PV) Annuity:** - Amount $A_{PV} = 3000$ - Interest rate = 8% annually, compounded quarterly, so $i = 0.08/4 = 0.02$ - Term = 6 years, so $n = 6 \times 4 = 24$ - Payment interval monthly, so effective monthly interest: $$i_{month} = (1 + 0.02)^{1/3} - 1 = 0.0066$$ - Total months $n = 6 \times 12 = 72$ Calculate $P$: $$P = \frac{A_{PV} \times i_{month}}{1 - (1 + i_{month})^{-n}} = \frac{3000 \times 0.0066}{1 - (1.0066)^{-72}}$$ Calculate denominator: $$(1.0066)^{-72} = \frac{1}{(1.0066)^{72}} = \frac{1}{e^{72 \times \ln(1.0066)}} = \frac{1}{e^{0.47}} = 0.625$$ So denominator = $1 - 0.625 = 0.375$ Calculate $P$: $$P = \frac{19.8}{0.375} = 52.8$$ Given payment is 19.76, which is too low. 4. **Summary of Corrected Periodic Payments:** - Simple (FV): $P = 465.9$ - Simple (PV): $P = 748.58$ - General (A): $P = 734.9$ - General (A_PV): $P = 52.8$ These corrected payments reflect proper use of annuity formulas considering compounding and payment intervals. **Final answer:** The original periodic payments given are incorrect; the corrected payments are as above.