Problem 36 - Algebra and Trigonometry

Question

Find the nth term \(a_{n}\) of each geometric sequence. When given, \(r\) is the common ratio. $$ 4,1, \frac{1}{4}, \frac{1}{16}, \ldots $$

Step-by-Step Solution

Answer

\( a_n = 4^{2-n} \)

Step 1: Identify the First Term

The first term of the given geometric sequence is 4. So, we have \[ a_1 = 4 \].

Step 2: Identify the Common Ratio

To find the common ratio, divide the second term by the first term: \[ r = \frac{1}{4} \div 4 = \frac{1}{4} \times \frac{1}{4} = \frac{1}{4} \].

Step 3: Use the nth Term Formula

The nth term of a geometric sequence is given by the formula: \[ a_n = a_1 \cdot r^{n-1} \].

Step 4: Substitute the Values

Substitute the values of the first term (\(a_1\)) and the common ratio (\(r\)) into the nth term formula: \[ a_n = 4 \cdot \left( \frac{1}{4} \right)^{n-1} \].

Step 5: Simplify the Formula

Simplify the expression to get the nth term: \[ a_n = 4 \cdot \left( \frac{1}{4} \right)^{n-1} = 4 \cdot \frac{1}{4^{n-1}} = 4 \cdot 4^{-(n-1)} = 4^{1-(n-1)} = 4^{2-n} \].

Key Concepts

Geometric Sequence

A geometric sequence, or geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio. This type of sequence can grow very quickly or shrink rapidly depending on whether the common ratio is greater than or less than one.

Consider the sequence \( 4, 1, \frac{1}{4}, \frac{1}{16}, \ldots \). Each term is obtained by multiplying the previous term by \(\frac{1}{4}\), the common ratio.

Let's break down what happens in a geometric sequence:

• The sequence starts with an initial term, also called the first term \(a_1\) (which is 4 here).
• The common ratio \(r\) (which is \(\frac{1}{4}\) in this example) is used to generate subsequent terms.

Geometric sequences are often represented as \(a_1, a_1 \cdot r, a_1 \cdot r^2, a_1 \cdot r^3, \ldots\).

nth Term Formula

To find any term in a geometric sequence, you use the nth term formula. This powerful formula lets you calculate the value of any term without counting all previous terms.

The formula is: \[ a_n = a_1 \cdot r^{n-1} \]

• \(a_n\) represents the nth term.
• \(a_1\) is the first term of the sequence.
• \(r\) is the common ratio.
• \(n\) is the term number.

Using our sequence as an example, if you want to find the 5th term:
\[ a_5 = 4 \cdot \left( \frac{1}{4} \right)^{5-1} \] By simplifying, you get: \[ a_5 = 4 \cdot \left( \frac{1}{4} \right)^4 = 4 \cdot \frac{1}{256} = \frac{4}{256} = \frac{1}{64} \]

Common Ratio

The common ratio in a geometric sequence is the factor that we multiply by to get from one term to the next. It is central to defining and understanding the sequence.

To find the common ratio \(r\) of a sequence, we divide any term by the term preceding it. For the given sequence \(4, 1, \frac{1}{4}, \frac{1}{16}, \ldots\), the common ratio is:
\[ r = \frac{1}{4} \] You can verify it by noting:

• \( \frac{1}{4} \div 4 = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{4}\)

An important insight is that if \(r\) is greater than 1, the sequence grows exponentially. If \(r\) is between 0 and 1, the sequence decreases towards zero.

Sequence Simplification

After identifying key values in the sequence (i.e., the first term and the common ratio), you need to simplify the expression to make calculations simpler and more understandable.

Using the nth term formula:

• Start with the general form: \(a_n = 4 \cdot \left(\frac{1}{4}\right)^{n-1}\)
• Express \(\frac{1}{4}\) to a power: \( a_n = 4 \cdot \frac{1}{4^{n-1}}\)
• Convert \(\frac{1}{4^{n-1}}\) to \(4^{-(n-1)}\): \(a_n = 4 \cdot 4^{-(n-1)}\)
• Simplify further: \(a_n = 4^{1-(n-1)}\)
• Finally, obtain the simplified result: \(a_n = 4^{2-n}\)

This simplification process ensures the nth term is expressed in the most straightforward form.