Question

A professional football player signs a six-year 2.50 million dollar contract, with a guaranteed 30% increase per year if he makes the All Pro team. If the player plays out the contract and makes the All Pro team each year, find the nth term of the geometric sequence that generates his annual salary and approximate his salary during the last year of his contract. 9.28 million dollars 7.14 million dollars 12.1 million dollars 9 million dollars MEDALS GIVEN HELP PEASE

Answer 1

\[2.5*10^{6}(1.3)^{n-1}=salary\]

Answer 2

This helped me ton. Why do you use this equation? Were does the 10 come from?

Answer 3

\[10^{6}\] is 1,000,000 or one million.

So 2.5 times one million is 2.5 million. So 2.5 million is the same as: \[2.5 * 10^{6}\]

He makes 100% of his salary plus an additional raise of 30% for each year he makes the all pro team. So his new salary for the start of next year is 130% of his old salary.

Change the percent to decimal by dividing by one hundred so 130/100=1.3 So that's where the 1.3 comes from.

Answer 4

The exponent comes from multiplying 1.3 multiple times. Where n= the number of years

First year he earns 2.5 x 10^6 *(1) n=year 1

Note that 1.3^0=1 or we can write it as (1.3)^(1-1)


Second year he earns (2.5 x 10^6)(1.3) n= year 2

Note that 1.3^(2-1) or we can write as 1.3^1=1.3

Third year he earns (2.5 x 10^6)(1.3)(1.3) n=year 3

Note that 1.3^(3-1) or we can write as 1.3^2

So the general equation for the exponent is n-1, where n is the year

Answer 5

For Algebra two students, the 1.3 number is called the common ratio and is given by the formula:\[\frac{ a _{n+1} }{ a _{n} }=r\] and the nth term is figured using this formula:

\[a _{n}=a _{1}*r ^{n-1} \] for n greater than or equal to 1