Question

Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{e_{n}\right\\}=\left\\{7^{n / 4}\right\\} $$

Short answer

The sequence is geometric with a common ratio of \(7^{\frac{1}{4}}\). The first four terms are \(7^{\frac{1}{4}}, 7^{\frac{1}{2}}, 7^{\frac{3}{4}},\) and \(7\).

Step-by-step solution

Identify the general term of the sequence

The given sequence is \(\{e_{n}\}\) and its general term is defined as \(e_{n} = 7^{n / 4}\).

Determine the form of a geometric sequence

A sequence is geometric if the ratio of any term to its preceding term is constant. This ratio is called the common ratio \(r\).

Find the common ratio

To find the common ratio \(r\), calculate \(\frac{e_{n+1}}{e_{n}}\):\(r = \frac{e_{n+1}}{e_{n}} = \frac{7^{\frac{n+1}{4}}}{7^{\frac{n}{4}}}\).

Simplify the common ratio

Use properties of exponents to simplify: \(\frac{7^{\frac{n+1}{4}}}{7^{\frac{n}{4}}} = 7^{\frac{n+1}{4} - \frac{n}{4}} = 7^{\frac{1}{4}}\). Hence, the common ratio \(r\) is \(7^{\frac{1}{4}}\).

List the first four terms

Calculate the first four terms of the sequence:\(e_{1} = 7^{\frac{1}{4}}\)\(e_{2} = 7^{\frac{2}{4}} = 7^{\frac{1}{2}}\)\(e_{3} = 7^{\frac{3}{4}}\)\(e_{4} = 7^{1}\)

Key concepts

Common Ratio

The common ratio is a key concept in a geometric sequence. It is the constant factor between consecutive terms of the sequence. To determine if a sequence is geometric, you check if dividing any term by its previous term always gives the same result.
In our example with the sequence \( \{e_{n}\} = \{7^{n / 4}\}\), we calculated the common ratio as follows:

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