Question

?

In a company's first year in operation, it made an annual profit of
$247, 500. The profit of the company increased at a constant 16% per
year each year. How much total profit would the company make over
the course of its first 10 years of operation, to the nearest whole
number?
Sum of Geometric series

Answer 1

Total profit after 10 years =
`\$5,277,064`

Explanation:

Let
`a_n` represent the profit in the nth year

Then
`a_(n+1)` represents the profit in year
`n+`1

`\text{Common ratio } r = (a_(n+1))/(a_n)`

The sum of a geometric sequence is given by

`S_n = a_1 \cdot (1 - r^n)/(1-r)`

where

`a_1 =` first term

`r =` common ratio

`n =` number of terms

Calculation of r

To calculate r we see that the profit increases by 16% every year

16% = 16/100 = 0.16

If profit increases by 0.16, then next year's profit
= this year's profit(1 + 0.16)

= this year's profit x 1.16

r = 1.16 the ratio of a term to the previous term

In this problem we are given the first term as

`a_1 = 247,500`
`\text{ = profit in first year}`

`n = 10 =` number of years

`r = 1.16`

Plugging these values into equation [1] for the sum we get

`\begin{aligned}S_n &= a_1 \cdot (1 - r^n)/(1-r)\\\\&= 247500 \cdot (1-1.16^(10))/(1-1.16)\\\\& = 247500 \cdot (-3.41143)/(-0.16)\\& = 247500 \cdot 21.3215\\& = 5277064\end{aligned}`

Therefore the total profit after 10 years
= $5,277,064