# Question Video: Modeling with Geometric Sequences Mathematics • 10th Grade

A ball rebounds to π times its previous height after each bounce. It is observed to rebound to a tenth of its original height on the 6th bounce. What is the value of π? Round your answer to two decimal places.

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### Video Transcript

A ball rebounds to π times its previous height after each bounce. It is observed to rebound to a tenth of its original height on the sixth bounce. What is the value of π? Round your answer to two decimal places.

Letβs think about what is actually happening here. The ball is dropped. Each bounce is slightly shorter than the previous bounce. Letβs say the ball originally starts at a height of β units. Since the ball rebounds to π times its previous height after each bounce, after the first bounce, it will reach a height of π times β units. Then, after the second bounce, it will reach a height of π times πβ or π squared β. And so, if we take each of the heights of the ball, we see that weβre generating a geometric sequence. This is one where each term is found by multiplying the previous term by a fixed nonzero number. We call this the common ratio. And here we see the common ratio is simply π.

Now, in fact, thereβs a formula that we can use to find any term in a geometric sequence. For a geometric sequence with first term π and common ratio π, the πth term, π sub π, is π times π to the power of π minus one. Now, in this case, our first term is β, but our common ratio is π. So the πth term of our sequence is π sub π equals β times π to the power of π minus one.

Now, in fact, we are actually given some information about the height of the ball on the sixth bounce. We know that on the sixth bounce, π will be equal to seven since weβre finding the height of the ball after that bounce. And weβre told that it is observed to rebound to a tenth of its original height, so one-tenth of β. That means the seventh term in our sequence is one-tenth β. And we can substitute π equals seven and π sub π as one-tenth β into our formula. We get a tenth β equals β times π to the power of seven minus one or one-tenth β equals βπ to the sixth power.

Weβre going to divide through by β. And so we find that one-tenth is equal to π to the sixth power. We could use logarithms to solve this equation for π. Alternatively, if our calculators allow it, we simply find the sixth root of both sides. Now, normally, for even powers of π, weβd be looking to find both the positive and negative root. In this case though, we know that π has to be a positive number. So π is equal to the sixth root of one-tenth. That gives us 0.6812 and so on, which is 0.68 correct to two decimal places. And so we say that π is equal to 0.68.