a) A particular ball always rebounds 3/5 the distance it falls. If the ball is dropped from a height of 5 meters, how far will it travel before coming to a stop?
Explain the reasoning.
b) Prove that if a1, a2, a3, ... is a geometric sequence,then ln(a1), ln(a2), ln(a3)... is an arithmetic sequence.
a) To find out how far the ball will travel before coming to a stop, we need to determine the sum of an infinite geometric series.
The geometric series formula is given by S = a / (1 - r), where S is the sum of the series, a is the initial term, and r is the common ratio.
In this case, the ball rebounds 3/5 the distance it falls, so the common ratio (r) is 3/5. The ball is dropped from a height of 5 meters, which is the initial term (a).
Using the formula, S = 5 / (1 - 3/5), we simplify it as S = 5 / (2/5), which is equal to 5 * 5/2 = 25/2 = 12.5 meters.
Therefore, the ball will travel a total distance of 12.5 meters before coming to a stop.
b) To prove that if a1, a2, a3, ... is a geometric sequence, then ln(a1), ln(a2), ln(a3), ... is an arithmetic sequence, we need to show that the difference between consecutive terms in ln(a1), ln(a2), ln(a3), ... is constant.
Let's take two consecutive terms ln(a_i) and ln(a_{i+1}). We want to show that ln(a_{i+1}) - ln(a_i) is a constant value.
Using logarithmic properties, ln(a_{i+1}) - ln(a_i) can be rewritten as ln(a_{i+1}/a_i).
Since a_1, a_2, a_3, ... is a geometric sequence, we have a_{i+1}/a_i = r, where r is the common ratio of the sequence.
Therefore, ln(a_{i+1}/a_i) = ln(r).
As r is a constant value for the geometric sequence, ln(r) is also a constant value.
Thus, ln(a1), ln(a2), ln(a3), ... forms an arithmetic sequence with a common difference of ln(r).