Which exponential function can be used to show the geometric sequence 40, 160, 640, . . .?

Question

A) an = 4(40)n-1
B) an = 40(4)n-1
C) an = 40 + 4 (n-1)
D) an = 40(120)n-1

4. Every year that a car is on the road it loses value. If a car loses 15 percent of its value each year and its beginning value is $24,500, what will the value of the car be after five years?
A) $10,870.78
B) $49,278.25
C) $12,789.15
D) $9240.16

Answer 1

To answer both questions, let's analyze them one at a time.

1. Identifying the Exponential Function for the Geometric Sequence

We are given the geometric sequence: 40, 160, 640, ...

To identify the function that describes this sequence, we need to first find the common ratio (r).

• From 40 to 160: 160 / 40 = 4
• From 160 to 640: 640 / 160 = 4

The common ratio is 4, which means each term is multiplied by 4 to get the next term.

The first term (a₁) is 40. For a geometric sequence, the nth term can be described by the formula: \[ a_n = a_1 \cdot r^{(n-1)} \]

In this case:

• \( a_1 = 40 \)
• \( r = 4 \)

So the nth term is: \[ a_n = 40 \cdot 4^{(n-1)} \]

Thus, the correct answer is B) \( a_n = 40(4)^{n-1} \).

2. Calculating the Depreciation of the Car

The car loses 15% of its value each year. Thus, it retains 85% of its value (100% - 15% = 85% or 0.85 as a decimal).

The value after \( t \) years can be calculated using the formula: \[ V = P \cdot (1 - r)^t \] Where:

• \( V \) is the value after t years.
• \( P \) is the initial value ($24,500).
• \( r \) is the percentage of value lost (0.15).
• \( t \) is the number of years.

For this problem:

• \( P = 24,500 \)
• \( r = 0.15 \)
• \( t = 5 \)

Substituting these values into the formula: \[ V = 24,500 \cdot (0.85)^5 \]

Calculating \( (0.85)^5 \): \[ (0.85)^5 \approx 0.4437 \]

Now calculate \( V \): \[ V \approx 24,500 \cdot 0.4437 \approx 10,870.78 \]

Thus, the value of the car after five years is approximately $10,870.78, so the correct answer is A) $10,870.78.