Question - Calculating Accumulated Value of Ordinary Annuity with Quarterly Compounding
Solution:
The image shows a financial math problem that reads: "What is the accumulated value of deposits of $112,000 made at the end of every six months for three years if interest is at 8.48% compounded quarterly?"We are given the following details:- Regular deposits of $112,000 are made at the end of every six months (semiannually), which constitutes an ordinary annuity.- The total period is three years.- The nominal interest rate is 8.48% per annum, compounded quarterly.To calculate the accumulated value, we need to use the future value formula for an ordinary annuity, adjusting appropriately for the semiannual deposits and quarterly compounding.First, we calculate the effective interest rate per six months, since interest is compounded quarterly. The nominal annual rate is 8.48%, so the quarterly rate is 8.48%/4 = 2.12% per quarter.To get the effective semiannual rate, we use the formula for compound interest for two quarters (six months):\[ (1 + i)^n \]where $$ i $$ is the quarterly interest rate and $$ n $$ is the number of quarters in six months.The effective semiannual rate $$ i_{\text{semi}} $$ is\[ i_{\text{semi}} = (1 + 0.0212)^2 - 1 \]\[ i_{\text{semi}} = (1 + 0.0212)\times(1 + 0.0212) - 1 \]\[ i_{\text{semi}} = 1.04308544 - 1 \]\[ i_{\text{semi}} = 0.04308544 \]\[ i_{\text{semi}} \approx 0.0431 \text{ (or 4.31%)} \]Now that we have the effective semiannual rate, we can calculate the future value of the annuity over three years. For six deposits, the future value formula is:\[ FV = P \times \left( \frac{(1 + i)^n - 1}{i} \right) \]where:$$ P $$ = periodic payment (in this case, $112,000),$$ i $$ = effective interest rate per period (4.31% per six months),$$ n $$ = total number of periods (six periods for three years).Plugging in our values:\[ FV = \$112,000 \times \left( \frac{(1 + 0.0431)^6 - 1}{0.0431} \right) \]Next, we calculate each part in turn:\[ (1 + 0.0431)^6 = (1.0431)^6 \approx 1.2846348545 \]\[ (1.0431)^6 - 1 \approx 0.2846348545 \]\[ \frac{(1.0431)^6 - 1}{0.0431} \approx 6.6014822329 \]\[ FV = \$112,000 \times 6.6014822329 \]\[ FV \approx \$739,365.9304828 \]Thus, the accumulated value of the deposits is approximately $739,365.93.
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