Question

homework #3

1. a local club is building a scholarship fund that earns 3% interest per year. a student is 11 years old today, and the club will award the scholarship when the student turns 18. the club deposits $4,000 today, then deposits $1,000 one year from today. each year after that, the club increases its annual deposit by $500, making the final deposit on the students 18th birthday. determine: (1) the account balance immediately after the final deposit, and (2) the uniform annual equivalent of the entire deposit series over the saving period.

Explanation:

First, calculate the number of years from today to the student's 18th birthday: $18 - 11 = 7$ years. So there are 7 deposits in total.

Step1: List all deposit amounts

• Deposit 0 (today): $P_0 = \$4000$
• Deposit 1 (1 year from now): $P_1 = \$1000$
• Deposit 2: $P_2 = \$1000 + \$500 = \$1500$
• Deposit 3: $P_3 = \$1500 + \$500 = \$2000$
• Deposit 4: $P_4 = \$2000 + \$500 = \$2500$
• Deposit 5: $P_5 = \$2500 + \$500 = \$3000$
• Deposit 6 (final, 6 years from now/18th birthday): $P_6 = \$3000 + \$500 = \$3500$

Step2: Calculate future value of each deposit

Use future value formula $FV = P \times (1 + r)^n$, where $r = 0.03$, $n$ is years until final deposit.

• $FV_0 = 4000 \times (1+0.03)^6 = 4000 \times 1.194052 = \$4776.21$
• $FV_1 = 1000 \times (1+0.03)^5 = 1000 \times 1.159274 = \$1159.27$
• $FV_2 = 1500 \times (1+0.03)^4 = 1500 \times 1.125509 = \$1688.26$
• $FV_3 = 2000 \times (1+0.03)^3 = 2000 \times 1.092727 = \$2185.45$
• $FV_4 = 2500 \times (1+0.03)^2 = 2500 \times 1.0609 = \$2652.25$
• $FV_5 = 3000 \times (1+0.03)^1 = 3000 \times 1.03 = \$3090$
• $FV_6 = 3500 \times (1+0.03)^0 = \$3500$

Step3: Sum future values for part (1)

Add all $FV$ values:
$$4776.21 + 1159.27 + 1688.26 + 2185.45 + 2652.25 + 3090 + 3500$$
$$= 4776.21+1159.27=5935.48; 5935.48+1688.26=7623.74;$$
$$7623.74+2185.45=9809.19; 9809.19+2652.25=12461.44;$$
$$12461.44+3090=15551.44; 15551.44+3500=19051.44$$

Step4: Calculate present value of all deposits

Use present value formula $PV = P \times (1 + r)^{-n}$:

• $PV_0 = 4000$
• $PV_1 = 1000 \times (1.03)^{-1} = 970.87$
• $PV_2 = 1500 \times (1.03)^{-2} = 1500 \times 0.9426 = 1413.90$
• $PV_3 = 2000 \times (1.03)^{-3} = 2000 \times 0.9151 = 1830.20$
• $PV_4 = 2500 \times (1.03)^{-4} = 2500 \times 0.8885 = 2221.25$
• $PV_5 = 3000 \times (1.03)^{-5} = 3000 \times 0.8626 = 2587.80$
• $PV_6 = 3500 \times (1.03)^{-6} = 3500 \times 0.8375 = 2931.25$

Sum all PV values:
$$4000 + 970.87 + 1413.90 + 1830.20 + 2221.25 + 2587.80 + 2931.25 = 15955.27$$

Step5: Find uniform annual equivalent (A)

Use capital recovery formula $A = PV \times \frac{r(1+r)^n}{(1+r)^n - 1}$, where $n=6$ (annual deposits start at year 1, 6 annual payments)
$$A = 15955.27 \times \frac{0.03(1.03)^6}{(1.03)^6 - 1}$$
First calculate $(1.03)^6 = 1.194052$
$$\frac{0.03 \times 1.194052}{1.194052 - 1} = \frac{0.0358216}{0.194052} = 0.1846$$
$$A = 15955.27 \times 0.1846 \approx 2945.34$$

Answer:

(1) The account balance immediately after the final deposit is $\$19051.44$
(2) The uniform annual equivalent of the deposit series is $\$2945.34$