Use Table \( 12-1 \) of your text to find the future value of \( \$ 1,300 \) deposited at the BEGINNING of every three months, for 3 years if the bank pays \( 12 \% \) interest, compounded quarterly. Select one: a. \( \$ 18,789.65 \) b. \( \$ 19,955.20 \) c. \( \$ 19,003.13 \) d. \( \$ 20,830.22 \)

To find the future value of $1,300 deposited at the beginning of every three months for 3 years with an interest rate of 12% compounded quarterly, we can use the formula for compound interest: \[ FV = PV \times (1 + \frac{r}{n})^{nt} \] Where: - \( FV \) is the future value - \( PV \) is the present value (initial deposit) - \( r \) is the annual interest rate - \( n \) is the number of times interest is compounded per year - \( t \) is the time in years Given: - \( PV = \$1,300 \) - \( r = 12\% = 0.12 \) - \( n = 4 \) (compounded quarterly) - \( t = 3 \) years Substitute the values into the formula: \[ FV = 1300 \times (1 + \frac{0.12}{4})^{4 \times 3} \] \[ FV = 1300 \times (1 + 0.03)^{12} \] \[ FV = 1300 \times (1.03)^{12} \] Now, we can calculate the future value using the formula. Calculate the value by following steps: - step0: Calculate: \(1300\left(1+0.03\right)^{12}\) - step1: Add the numbers: \(1300\times 1.03^{12}\) - step2: Convert the expressions: \(1300\left(\frac{103}{100}\right)^{12}\) - step3: Simplify: \(1300\times \frac{103^{12}}{100^{12}}\) - step4: Rewrite the expression: \(100\times 13\times \frac{103^{12}}{100^{12}}\) - step5: Reduce the numbers: \(13\times \frac{103^{12}}{100^{11}}\) - step6: Multiply: \(\frac{13\times 103^{12}}{100^{11}}\) The future value of $1,300 deposited at the beginning of every three months for 3 years with an interest rate of 12% compounded quarterly is approximately $1,853.49. Therefore, the correct answer is not among the options provided. The future value is approximately $1,853.49, which is not listed in the options.