Find the nth term an of each geometric sequence. When given, r is the common ratio. a3=(1)/(3) ;

Find the nth term an of each geometric sequence. When given,...

Key Concepts

Geometric Sequence

A geometric sequence is a list of numbers where each term after the first is obtained by multiplying the previous term by a constant called the common ratio. This concept is fundamental in algebra and appears in various applications involving exponential growth or decay.

Common Ratio

The common ratio is the constant factor between consecutive terms of a geometric sequence. It is a key parameter that defines the sequence and can be calculated by dividing any term by the previous term, ensuring consistent proportionality throughout the sequence.

Nth Term Formula

The nth term of a geometric sequence is given by the formula a? = a? × r^(n-1), where a? is the first term and r is the common ratio. This generalized expression allows for the determination of any term within the sequence without listing all preceding terms.

Solving for Unknown Parameters

When specific terms in a geometric sequence are provided, such as the third and sixth terms, these values can be substituted into the nth term formula to create equations. Solving these equations simultaneously helps to identify the unknown parameters like the first term and the common ratio, thereby fully determining the sequence.

Transcript

00:01 Suppose we have a sequence where the third term is one -third and the sixth term is 1 over 81.

00:05 Let's find what the generic nth term will look like.

00:08 According to the book, we have this formula where the nth term, a -n, is equal to the first term, a1, multiplied by the common ratio r raised to the n -minus -1 power.

00:18 So if we want to figure out what the nth term looks like, we need to figure out a -1 and r.

00:23 Let's get started with r.

00:25 So think about the definition of the common ratio.

00:27 If you multiply one term, that is, let's say the first term, a1, by r, you will get the second term, a2.

00:36 And then if you multiply a2 by r, you will get the third term.

00:42 So, with this in mind, you could, for example, substitute in and find that a1 times r times r will just give you directly the third term.

00:56 That is if you multiply both sides of this equation by r, you will get a1 times r squared on the left and a3 on the right with the substitution.

01:08 So now, with this in mind, let's try something a little bit more practical for this problem.

01:13 That is, suppose we took the term a3 and multiplied it by r cubed.

01:18 What would that give us? well, let's see.

01:23 So if we have a3 times r cubed, a3 times u.

01:28 Would give us a 4, another r, a 5, and then another r, what we have, would give us a 6, which we're given.

01:36 So we can substitute what we have, 1 3rd times r cubed equals 1 over 81...

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