## Precalculus (6th Edition) Blitzer

At the end of the year, the value of the car is $\dfrac{3}{4}$ of the value of the previous year.
Here, we have $a_5=25,000 (\dfrac{3}{4})^5 \approx 5933$ and $a_{n+1}=25,000 (\dfrac{3}{4})^{n+1}$ or, $a_{n+1}=25,000 (\dfrac{3}{4})^{n} \cdot (3/4) \approx 5933$ Thus, $a_{n+1}=a_n (\dfrac{3}{4})$ This implies that at the end of the year the value of the car is $\dfrac{3}{4}$ of the value of the previous year.