Suppose that $780 is deposited at the end of every year into an account paying interest of 6% per year. At the end of fifteen (15) years the account will be worth approximately $.

The account will be worth approximately $1,869

Explanation:

First of all, note that 6% of an amount = 6/100 × the amount = 0.06×amount.

Next let us calculate the amounts gotten for the first 4 years, and establish a pattern that will will us for the remaining 11 years.

1st year total= deposit + (0.06×deposit) = 780 + (0.06 × 780)

= 780 + 46.8 = $826.8

2nd year total = 826.8 + (0.06 × 826.8) = $876.408

3rd year total = 876.408 + (0.06 × 876.408) = $928.992

4th year total = 928.992 + (0.06 × 928.992) = $984.732

Now, if we observe the total amounts as the year progresses, we notice that the next year increase by a certain constant factor which is 1.06; this is determined by dividing the amount in a year by the amount in the previous year. It is shown below;

Year 2 ÷ Year 1 = 876.408 ÷ 826.8 = 1.06

year 3 ÷ year 2 = 928.992 ÷ 876.408 = 1.06

year 4 ÷ year 3 = 984.732 ÷ 928.992 = 1.06

Now, to determine the amount in the next year, we will multiply the amount in the previous year by 1.06 (common increasing factor)

Year 5 = year 4 × 1.06 = 984.732 × 1.06 = $1,046.816

year 6 = 1046.816 × 1.06 = $1,106.445

year 7 = 1106.445 × 1.06 = $1,172.832

year 8 = 1172.832 × 1.06 = $1,243.202

year 9 = 1243.202 × 1.06 = $1,317.794

year 10 = 1317.794 × 1.06 = $1,396.862

year 11 = 1396.862 × 1.06 = $1,480.674

year 12 = 1480.674 × 1.06 = $1,569.514

year 13 = 1569.514 × 1.06 = $1,663.684

year 14 = 1663.685 × 1.06 = $1,763.506

Year 15 = 1763.506 × 1.06 = $1,869.316 which is approximately $1,869

Alternatively, you can count how many 1.06s are there from year 5 to year 15, and the answer is 11. then you can raise 1.06 to a power of 11 as shown

then multiply the amount in year 4 by 1.8983

= 984.732 × 1.8983 = $1,869.316. = approx. $1869

This second method is easier, but I wanted you to see what is going on that is why i did the details in the first method

answer; tech (a) is correct;