Chapter 12.6, Problem 57SR

To determine

To find: Salary after four consecutive $4\%$ increases.

Answer to Problem 57SR

Salary after four consecutive $4\%$ increases is $\$46,800$

Explanation of Solution

Given information:

Current salary of Phillip is

  $a_0=\$40,000$

He got four concecutive $4\%$ increases.

Given

  $a_0=\$40,000$

Every year salary increases by $4\%$ .

It means

Total salary after one year

Will be

Salary $=40000+\frac{4}{100}\times 40000$

  $\Rightarrow$ Salary $=40000\left(1+\frac{4}{100}\right)$

  $\Rightarrow$ Salary $=40000\left(1+0.04\right)$

  $\Rightarrow$ Salary $=40000\left(1.04\right)$

Therefore,

Salary after one year wil be

  $\Rightarrow a_1=40000{\left(1.04\right)}^1$

Salary after two years will be

Salary $=40000{\left(1.04\right)}^1+\frac{4}{100}\times 40000{\left(1.04\right)}^1$

  $\Rightarrow$ Amount $=40000{\left(1.04\right)}^1\left(1+\frac{4}{100}\right)$

  $\Rightarrow$ Amount $=40000{\left(1.04\right)}^1\left(1+0.04\right)$

  $\Rightarrow$ Amount $=40000{\left(1.04\right)}^1\left(1.04\right)$

  $\Rightarrow$ Amount $=40000{\left(1.04\right)}^2$

Therefore,

Salary after two years is

  $\begin{array}{l}\Rightarrow a_2=40000{\left(1.04\right)}^2\\ \Rightarrow a_2=a_0{\left(1.04\right)}^2\end{array}$

Therefore,

On comparing $a_1,a_2$

We can write a general equation that is

  $\Rightarrow a_n=a_0{\left(1.04\right)}^n$

Therefore,

Recursive equation for the Salary after $n$ years is

  $a_n=a_0{\left(1.04\right)}^n$

We need to find the salary after foure consecutive $4\%$ increases.

It means

Substitute $n=4$

We get

  $\begin{array}{l}{a}_n=a_0{\left(1.04\right)}^n\\ \Rightarrow a_4=40000{\left(1.04\right)}^4\\ \Rightarrow a_4=40000\left(1.17\right)\\ \Rightarrow a_4=46800\end{array}$

Therefore,

Salary after four consecutive $4\%$ increases is $\$46,800$