Question

Are the sequences in Exercises \(1-6\) geometric? For those that are, give a formula for the \(n^{\text {th }}\) term. $$ 8,4,2,1, \frac{1}{2}, \frac{1}{4}, \ldots $$

Short answer

Answer: The formula for the nth term of this geometric sequence is \(a_n = 8 \times \left(\frac{1}{2}\right)^{(n-1)}\).

Step-by-step solution

Identify the sequence terms

The sequence given is: $$ 8, 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots $$

Check for a common ratio between consecutive terms

The common ratio is found by dividing each term by the previous term. Let's check the common ratio for the given sequence: $$ \frac{4}{8} = \frac{1}{2} $$ $$ \frac{2}{4} = \frac{1}{2} $$ $$ \frac{1}{2} = \frac{1}{2} $$ $$ \frac{\frac{1}{2}}{1} = \frac{1}{2} $$ $$ \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} $$ The common ratio between consecutive terms is \(\frac{1}{2}\).

Confirm that the sequence is geometric

Since we found a common ratio between all consecutive terms in the sequence, we can confirm that the sequence is geometric.

Find the formula for the nth term

The formula for the nth term of a geometric sequence is given by: $$ a_n = a_1 \times r^{(n-1)} $$ where \(a_n\) is the nth term, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. In this sequence, the first term \(a_1 = 8\), and the common ratio \(r = \frac{1}{2}\). Plug in these values and we get: $$ a_n = 8 \times \left(\frac{1}{2}\right)^{(n-1)} $$ The formula for the nth term of this geometric sequence is: $$ a_n = 8 \times \left(\frac{1}{2}\right)^{(n-1)} $$

Key concepts

Common Ratio in Geometric Sequences

To determine if a sequence is geometric, we need to look for something called the common ratio. The common ratio is the number that we multiply by to get from one term to the next in a sequence. You find it by taking a term in the sequence and dividing it by the previous term. For instance:

• If our sequence starts with 8, 4, 2, 1, \( \frac{1}{2} \), \( \frac{1}{4} \), and so on, we calculate \( \frac{4}{8} \) to see if we can use the same common ratio throughout.
• In this sequence, \( \frac{4}{8}, \frac{2}{4}, \frac{1}{2}, \frac{1}{1}, \frac{\frac{1}{4}}{\frac{1}{2}} \) all yield \( \frac{1}{2} \).

Every term divided by the preceding term results in \( \frac{1}{2} \). This consistent ratio throughout the sequence confirms that the sequence is geometric. Coming across a constant common ratio is what transforms a random sequence into a pattern-rich geometric sequence.

Formula for the nth Term of a Geometric Sequence

Once we have identified the common ratio for a geometric sequence, we can build a formula that makes it easy to find any term in the sequence without having to compute each previous term. This is the nifty part of understanding sequences: formulas. The formula for the nth term \( a_n \) of a geometric sequence is:\[a_n = a_1 \times r^{(n-1)}\]- \( a_n \) represents the term you want to find. - \( a_1 \) is the first term of the sequence. - \( r \) is the common ratio you found.- \( n \) is the position of the term in the sequence. In our case:- First term \( a_1 = 8 \)- Common ratio \( r = \frac{1}{2} \)Plug these into the formula:\[a_n = 8 \times \left(\frac{1}{2}\right)^{(n-1)}\]This formula allows you to calculate any term in this geometric sequence, whether it’s the 10th term or the 100th term, quickly and accurately.

Geometric Series Formula

When you're working with a geometric sequence, it often helps to consider not only the individual terms but the sum of several terms, which is called a geometric series. The formula for the sum of the first \( n \) terms \( S_n \) of a geometric sequence lets us add up these terms quite efficiently:\[S_n = a_1 \times \frac{1 - r^n}{1 - r}\]This is useful when you want to know how much you have in total after adding multiple terms of your sequence. - \( S_n \): the sum of the first \( n \) terms- \( a_1 \): the first term- \( r \): the common ratio- \( n \): the number of terms you are summing up For example, summing the first 5 terms of our sequence where \( a_1 = 8 \), \( r = \frac{1}{2} \), and \( n = 5 \):\[S_5 = 8 \times \frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}} \]You get a swift and easy way to calculate the sum without manually adding each term individually. Deconstruct the pattern into something you can add up quickly! This is what makes the study of sequences and series powerful in mathematics.