Question

a. use the appropriate formula to determine the periodic deposit.
b. how much of the financial goal comes from deposits and how much comes from interest?

periodic deposit	rate	time	financial goal

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a. the periodic deposit is $\square
(do not round until the final answer. then round up to the nearest dollar.)

Explanation:

Step1: Identify the formula for periodic deposit (annuity)

The formula for the future value of an ordinary annuity is \( FV = P \times \frac{(1 + r)^n - 1}{r} \), where \( FV \) is the future value (financial goal), \( P \) is the periodic deposit, \( r \) is the interest rate per period, and \( n \) is the number of periods.

First, find the interest rate per period and the number of periods.
- The annual interest rate \( i = 3.5\% = 0.035 \). Since it's compounded quarterly, the rate per period \( r=\frac{0.035}{4}=0.00875 \).
- The time is 6 years, and since compounding is quarterly (every 3 months), the number of periods \( n = 6\times4 = 24 \).
- The future value \( FV = 15000 \).

Step2: Rearrange the formula to solve for \( P \)

From \( FV = P \times \frac{(1 + r)^n - 1}{r} \), we can solve for \( P \):
\( P=\frac{FV\times r}{(1 + r)^n - 1} \)

Substitute the values:
\( r = 0.00875 \), \( n = 24 \), \( FV = 15000 \)

First, calculate \( (1 + r)^n=(1 + 0.00875)^{24} \)
\( (1.00875)^{24}\approx1.23299 \) (using a calculator)

Then, calculate the numerator: \( FV\times r = 15000\times0.00875 = 131.25 \)

Calculate the denominator: \( (1.00875)^{24}- 1\approx1.23299 - 1 = 0.23299 \)

Now, \( P=\frac{131.25}{0.23299}\approx563.32 \)

Since we need to round up to the nearest dollar, \( P = 564 \) (Wait, let's check the calculation again. Maybe I made a mistake in the exponent calculation. Let's recalculate \( (1.00875)^{24} \) more accurately.

Using the formula \( a^b=e^{b\ln a} \), \( \ln(1.00875)\approx0.008715 \), \( 24\times0.008715\approx0.20916 \), \( e^{0.20916}\approx1.2328 \)

Then \( (1.00875)^{24}-1\approx1.2328 - 1 = 0.2328 \)

\( 15000\times0.00875 = 131.25 \)

\( P=\frac{131.25}{0.2328}\approx563.79 \). Rounding up to the nearest dollar gives \( 564 \)? Wait, no, maybe my initial approach is wrong? Wait, no, let's check the formula again. Wait, the formula for ordinary annuity (payments at the end of the period) is correct. Wait, maybe I miscalculated the exponent. Let's use a calculator for \( (1.00875)^{24} \):

\( 1.00875^{1}=1.00875 \)

\( 1.00875^{2}=1.00875\times1.00875 = 1.0175765625 \)

\( 1.00875^{4}=(1.0175765625)^{2}\approx1.03548 \)

\( 1.00875^{8}=(1.03548)^{2}\approx1.07227 \)

\( 1.00875^{16}=(1.07227)^{2}\approx1.1497 \)

Then \( 1.00875^{24}=1.00875^{16}\times1.00875^{8}\approx1.1497\times1.07227\approx1.2329 \)

So denominator is \( 1.2329 - 1 = 0.2329 \)

\( 15000\times0.00875 = 131.25 \)

\( 131.25\div0.2329\approx563.5 \). Rounding up to the nearest dollar, we get \( 564 \)? Wait, but maybe the correct calculation is as follows:

Wait, let's use the formula step by step with more precise calculation.

\( P=\frac{15000\times0.00875}{(1 + 0.00875)^{24}-1} \)

Calculate \( (1.00875)^{24} \):

Using a calculator, \( 1.00875^{24}=e^{24\ln(1.00875)} \)

\( \ln(1.00875)=0.0087153 \)

\( 24\times0.0087153 = 0.209167 \)

\( e^{0.209167}\approx1.23289 \)

So \( (1.00875)^{24}-1 = 0.23289 \)

Numerator: \( 15000\times0.00875 = 131.25 \)

\( P = 131.25\div0.23289\approx563.5 \). Since we need to round up to the nearest dollar, the periodic deposit is \( \$564 \)? Wait, maybe I made a mistake. Wait, let's check with an online annuity calculator.

Alternatively, maybe the formula is for future value of an ordinary annuity, and the calculation is correct. So the periodic deposit is approximately \( \$563 \) when rounded to the nearest dollar? Wait, no, let's recalculate:

\( 131.25\div0.23289 = 563.5 \), so when we round up, it's \( 564 \). But let's check the exact value.

Wait, maybe the initial formula is correct. So the periodic deposit is approximately \( \$563 \) or \( \$564 \)? Wait, let's do the division again: \( 131.25\div0.23289\approx563.5 \), so rounding up to the nearest dollar is \( 564 \).

Answer:

\( \$564 \) (Wait, but let's check with the formula again. Maybe I made a mistake in the number of periods. Wait, 6 years, quarterly, so 6*4=24 periods. Rate per period 3.5%/4=0.875%=0.00875. Future value 15000.

Using the formula \( P = \frac{FV \times r}{(1 + r)^n - 1} \)

\( r = 0.00875 \), \( n = 24 \), \( FV = 15000 \)

\( (1 + 0.00875)^{24} = 1.00875^{24} \approx 1.23289 \)

\( (1.23289 - 1) = 0.23289 \)

\( 15000 * 0.00875 = 131.25 \)

\( 131.25 / 0.23289 ≈ 563.5 \), so rounding up to the nearest dollar is \( 564 \). So the periodic deposit is \( \$564 \).)