Serenity deposits dollar sign, 1, comma, 000$1,000 every year into an account earning an annual interest rate of 6.5%, compounded annually. How many years would it be until Serenity had dollar sign, 25, comma, 000$25,000 in the account, to the nearest tenth of a year? Use the following formula to determine your answer.

Question

A, equals, d, left bracket, start fraction, left bracket, 1, plus, i, right bracket, to the power n , minus, 1, divided by, i, end fraction, right bracket
A=d(
i
(1+i)
n
−1
​
)
A, equalsA= the future value of the account after n periods
d, equalsd= the amount invested at the end of each period
i, equalsi= the interest rate per period
n, equalsn= the number of periods
the answer is not 15.7

Answer 1

To solve for the number of years until Serenity has $25,000 in the account, we can plug in the given values into the formula and solve for n:

A = $25,000
d = $1,000
i = 6.5% = 0.065 (decimal form)

$25,000 = $1,000 * ((1 + 0.065)^n - 1) / 0.065

$25,000 = $1,000 * ((1.065)^n - 1) / 0.065

$25 = (1.065)^n - 1

26 = (1.065)^n

Take the natural log of both sides to solve for n:

ln(26) = n * ln(1.065)

n = ln(26) / ln(1.065)

n ≈ 14.8

Therefore, it would take approximately 14.8 years until Serenity has $25,000 in the account.