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You want to purchase a new car in 6 years and expect the car to cost $10,000. Your bank offers a plan with a guaranteed APR of 6.5% if you make regular monthly deposits. How much should you deposit each month to end up with $10,000 in 6 years? You should invest $◻ each month. (Round the final answer to the nearest cent as needed. Round all intermediate values to seven decimal places as needed.)

You want to purchase a new car in 66 years and expect the car to cost $10,000 \$ 10,000 . Your bank offers a plan with a guaranteed APR of 6.5% 6.5 \% if you make regular monthly deposits. How much should you deposit each month to end up with $10,000 \$ 10,000 in 66 years?\newlineYou should invest $□ \$ \square each month.\newline(Round the final answer to the nearest cent as needed. Round all intermediate values to seven decimal places as needed.)

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Q. You want to purchase a new car in 66 years and expect the car to cost $10,000 \$ 10,000 . Your bank offers a plan with a guaranteed APR of 6.5% 6.5 \% if you make regular monthly deposits. How much should you deposit each month to end up with $10,000 \$ 10,000 in 66 years?\newlineYou should invest $□ \$ \square each month.\newline(Round the final answer to the nearest cent as needed. Round all intermediate values to seven decimal places as needed.)
  1. Identify Formula: Identify the formula to calculate the monthly deposit needed to reach a future value with regular deposits at a given interest rate.\newlineThe formula for the future value of a series of equal monthly deposits (ordinary annuity) is:\newlineFV=P×[(1+r)nt−1]/rFV = P \times \left[(1 + r)^{nt} - 1\right] / r\newlineWhere:\newlineFVFV = future value\newlinePP = monthly deposit\newlinerr = monthly interest rate\newlinenn = number of times the interest is compounded per year\newlinett = number of years\newlineWe need to solve for PP.
  2. Convert APR to Monthly Rate: Convert the annual percentage rate (APR) to a monthly interest rate.\newlineAPR = 6.5%6.5\% or 0.0650.065 as a decimal.\newlineMonthly interest rate (rr) = APR/12\text{APR} / 12\newliner=0.065/12r = 0.065 / 12\newliner≈0.0054166667r \approx 0.0054166667 (rounded to seven decimal places)
  3. Determine Compounding Frequency: Determine the number of times the interest is compounded per year nn and the total number of years tt. Since the deposits are monthly, the interest is compounded monthly, so n=12n = 12. The time period is 66 years, so t=6t = 6.
  4. Substitute Values and Solve: Substitute the values into the formula and solve for PP.FV=$10,000FV = \$10,000r=0.0054166667r = 0.0054166667n=12n = 12t=6t = 6Substitute these values into the formula:$10,000=P×[(1+0.0054166667)(12×6)−1]/0.0054166667\$10,000 = P \times [(1 + 0.0054166667)^{(12\times6)} - 1] / 0.0054166667
  5. Calculate Value Inside Brackets: Calculate the value inside the brackets.\newline(1+0.0054166667)(12∗6)−1(1 + 0.0054166667)^{(12*6)} - 1\newline= (1+0.0054166667)72−1(1 + 0.0054166667)^{72} - 1\newline= (1.0054166667)72−1(1.0054166667)^{72} - 1\newline≈1.489856259−1\approx 1.489856259 - 1\newline≈0.489856259\approx 0.489856259 (rounded to seven decimal places)
  6. Divide Future Value to Find P: Divide the future value by the result inside the brackets to find P.\newlineP=extextdollar10,0000.489856259P = \frac{ ext{ extdollar}10,000}{0.489856259}\newlineP≈extextdollar20,415.5810.489856259P \approx \frac{ ext{ extdollar}20,415.581}{0.489856259}\newlineP≈extextdollar20415.5810.489856259P \approx \frac{ ext{ extdollar}20415.581}{0.489856259}\newlineP≈extextdollar41.681P \approx ext{ extdollar}41.681 (rounded to the nearest cent)

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