Question

De Shawn is in his fifth year of employment as a patrol officer for the Metro Police. His salary for his first year of employment was $47,000. Each year after the first, his salary increased by 4% of his salary the previous year.

To obtain the sum of DeShawn's salaries for the first five years of service, we can use f(x)=47,000(1.04)^x , where x gives the salary x years after his first year.

Salary after 5 years = f(0) + f(1) + f(2) + f(3) + f(4) +f(5) = $245,567.

If DeShawn continues his employment at the same rate of increase in yearly salary, for which year will the sum of his salaries first exceed $1,000,000?

(The answer should be 16, but I don't know how to get it.)

Answer 1

Answer:

After 16 years De Shawn salary will exceed $1000000.

Step-by-step explanation:

First year salary of De Shawn as a petrol officer was $47000.

Each year increase in his salary was 4%.

Second year his salary will be = ($47000 + 4% of $47000) = $48880

Third year the salary will be = ($48880 + 4% of 48880) = $50835.20

Then every year his salary will be $47000, $48880, $50835...........

The sequence formed will be a geometric sequence having common ratio 'r' = (1+0.04) = 1.04

First term of this sequence = $47000

Since sum of a geometric sequence is represented by

$S_{n}=\frac{a(r^{x-1})}{r-1}$

We have to find the value of x (Number of years) after which the salary will be = $1000000

$1000000=\frac{4700(1.04^{x-1})}{1.04-1}$

$\frac{1000000}{47000}=\frac{1.04^{x}-1}{0.04}$

21.2766×0.04 + 1 = $(1.04)^{x}$

1.851 = $(1.04)^{x}$

By taking log on both the sides

log(1.851) = $log(1.04)^{x}$

0.2764 = xlog(1.04)

x = $\frac{0.2764}{0.017}$

x = 16.26 years

Therefore, after 16 year De Shawn's salary will exceed $1000000.