Problem 15 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

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 1: Identify the general term of the sequence

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

Step 2: 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 \(r\).

Step 3: Find the common ratio

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

Step 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 \(7^{\frac{1}{4}}\).

Step 5: 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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