Gilberto opened a savings account for his daughter and deposited​ $1200 on the day she was born. Each year on her​ birthday, he deposited another ​$1200. If the account pays ​9.5% ​interest, compounded​ annually, how much is in the account at the end of the day on her 10th ​birthday?

Answer 1

We can use the formula for compound interest:

A = P*(1+r/n)^(n*t)

where
A = the amount at the end of the investment period
P = the principal amount (the initial deposit)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years

In this case, P = $1200, r = 0.095, n = 1 (compounded annually), and t = 10.

For the first deposit, the amount after 1 year is:

A1 = 1200*(1+0.095/1)^(1*1) = $1314

For the second deposit, the amount after 2 years is:

A2 = (1200+1314)*(1+0.095/1)^(1*1) = $2871.30

For the third deposit, the amount after 3 years is:

A3 = (1200+1314+2871.30)*(1+0.095/1)^(1*1) = $4452.37

And so on, until we reach the 10th deposit and the amount after 10 years:

A10 = (1200+1314+2871.30+...+12207.63)*(1+0.095/1)^(1*10) = $29,992.60

Therefore, the amount in the account at the end of the day on her 10th birthday is $29,992.60.