What annual rate of interest, compounded annually, will double an investment in 7 years? Given that 21/7=1.104090.




We are given that an investment doubles in 7 years and we need to find the annual rate of interest that causes this doubling. We are also given that 21/7 is equal to 1.104090.


To find the annual rate of interest, we can use the compound interest formula:

A = P(1 + r/n)^(nt)

Where:
A = final amount (double the initial investment)
P = principal amount (initial investment)
r = annual interest rate (to be determined)
n = number of times interest is compounded per year (annually in this case)
t = number of years (7 in this case)

In our case, the final amount is twice the principal amount:

2P = P(1 + r/1)^(1*7)


We can simplify the equation:

2 = (1 + r)^7

Taking the seventh root of both sides:

∛2 = 1 + r

Subtracting 1 from both sides:

∛2 - 1 = r

Using the value of 21/7 = 1.104090, we can calculate the annual interest rate:

r = ∛2 - 1
r = 1.104090 - 1
r = 0.104090

Therefore, the annual rate of interest, compounded annually, that doubles an investment in 7 years is approximately 0.104090 or 10.4090%.


The annual rate of interest, compounded annually, that doubles an investment in 7 years is approximately 10.4090%.