3. 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

3. To determine which exponential function represents the geometric sequence 40, 160, 640, ..., we first identify the common ratio. In this case:

• From 40 to 160, the common ratio is \( \frac{160}{40} = 4 \).
• From 160 to 640, the common ratio is \( \frac{640}{160} = 4 \).

This confirms that the common ratio \( r \) is 4. The first term \( a \) of the sequence is 40. The general form of a geometric sequence is given by:

\[ a_n = a \cdot r^{n-1} \]

Thus, substituting in the values:

\[ a_n = 40 \cdot 4^{n-1} \]

The correct option that matches this is:

B) \( a_n = 40(4)^{n-1} \)

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4. To find the value of the car after five years, we need to calculate the depreciation using the formula for exponential decay:

\[ V = P (1 - r)^t \]

Where:

• \( P = 24,500 \) (initial value)
• \( r = 0.15 \) (15% depreciation rate)
• \( t = 5 \) (number of years)

Plugging in the values:

\[ V = 24500 \times (1 - 0.15)^5 \] \[ V = 24500 \times (0.85)^5 \] \[ V = 24500 \times 0.4437053125 \quad \text{(approximately)} \] \[ V \approx 10870.78 \]

Thus, the value of the car after five years is approximately $10,870.78.

The correct option is:

A) $10,870.78