When a rubber ball is dropped it bounces back up to a height 3/4 as high as its previous height. You drop this ball from a height of 16 feet and measure its height after each bounce. Which formula represents a geometric sequence that describes the height of the ball after the number of bounces n ? (1 point) Responses an=16(34)n−1 A sub n is equal to 16 times 3 fourths raised to the n minus 1 power an=34(16)n−1 A sub n is equal to 3 fourths times 16 raised to the n minus 1 power an=16+(34)n−1 A sub n is equal to 16 plus 3 fourths raised to the n minus 1 power an=(n−1)(34)(16) A sub n is equal to open paren n minus 1 close paren times 3 fourths times 16

1 answer

To represent the height of the rubber ball after \( n \) bounces, we can use the formula for a geometric sequence. The initial height from which the ball is dropped is 16 feet, and after each bounce, the ball reaches \( \frac{3}{4} \) of the height of its previous bounce.

Let’s define the sequence:

  1. The height before any bounce (when \( n = 0 \)) is \( a_0 = 16 \) feet.
  2. After the first bounce (when \( n = 1 \)), the height is \( a_1 = 16 \times \frac{3}{4} \).
  3. After the second bounce (when \( n = 2 \)), the height is \( a_2 = 16 \times \left(\frac{3}{4}\right)^2 \).
  4. In general, after \( n \) bounces, the height can be expressed as:

\[ a_n = 16 \times \left(\frac{3}{4}\right)^n. \]

Notice that in this case, we want to find the height after \( n \) bounces, so the formula from above can be adjusted slightly. Since we want \( n \) to represent the number of bounces (where \( n \) starts counting from 0 for the drop itself), if we set \( n \) to start from 1 instead, we can write:

\[ a_n = 16 \times \left(\frac{3}{4}\right)^{n-1}. \]

Thus, the correct formula representing the height of the ball after \( n \) bounces is given by:

\[ a_n = 16 \times \left(\frac{3}{4}\right)^{n-1}. \]

So the answer is:

\( a_n = 16 \left( \frac{3}{4} \right)^{n-1} \).