Problem 19 - Algebra and Trigonometry

Question

Find the first five terms of the sequence and determine if it is geometric. If it is geometric, find the common ratio and express the \(n\)th term of the sequence in the standard form \(a_{n}=a r^{n-1} .\) $$ a_{n}=\frac{1}{4^{n}} $$

Step-by-Step Solution

Answer

The sequence is geometric with a common ratio of \( \frac{1}{4} \).

Step 1: Find the First Term

To find the first term of the sequence, substitute \( n = 1 \) into the formula \( a_n = \frac{1}{4^n} \). \[ a_1 = \frac{1}{4^1} = \frac{1}{4} \] The first term \( a_1 \) is \( \frac{1}{4} \).

Step 2: Find the Second Term

To find the second term, substitute \( n = 2 \) into the formula.\[ a_2 = \frac{1}{4^2} = \frac{1}{16} \] The second term \( a_2 \) is \( \frac{1}{16} \).

Step 3: Find the Third Term

To find the third term, substitute \( n = 3 \).\[ a_3 = \frac{1}{4^3} = \frac{1}{64} \] The third term \( a_3 \) is \( \frac{1}{64} \).

Step 4: Find the Fourth Term

To find the fourth term, substitute \( n = 4 \).\[ a_4 = \frac{1}{4^4} = \frac{1}{256} \] The fourth term \( a_4 \) is \( \frac{1}{256} \).

Step 5: Find the Fifth Term

To find the fifth term, substitute \( n = 5 \).\[ a_5 = \frac{1}{4^5} = \frac{1}{1024} \] The fifth term \( a_5 \) is \( \frac{1}{1024} \).

Step 6: Check if the Sequence is Geometric

A sequence is geometric if the ratio \( r \) between any two consecutive terms \( \frac{a_{n+1}}{a_n} \) is constant. Let's calculate the ratios:\[ r = \frac{a_2}{a_1} = \frac{\frac{1}{16}}{\frac{1}{4}} = \frac{1}{16} \times 4 = \frac{1}{4} \ r = \frac{a_3}{a_2} = \frac{\frac{1}{64}}{\frac{1}{16}} = \frac{1}{64} \times 16 = \frac{1}{4} \ r = \frac{a_4}{a_3} = \frac{\frac{1}{256}}{\frac{1}{64}} = \frac{1}{256} \times 64 = \frac{1}{4} \ r = \frac{a_5}{a_4} = \frac{\frac{1}{1024}}{\frac{1}{256}} = \frac{1}{1024} \times 256 = \frac{1}{4}\]Since the ratio \( r \) is consistent \( \frac{1}{4} \), the sequence is geometric.

Step 7: Find the Common Ratio and General Formula

The common ratio \( r \) is \( \frac{1}{4} \). The first term \( a_1 = \frac{1}{4} \). Thus, the general formula for the \( n \)th term of the sequence in the standard form \( a_n = a r^{n-1} \) is:\[ a_n = \frac{1}{4} \left( \frac{1}{4} \right)^{n-1} = \frac{1}{4^n} \]

Key Concepts

Common Ratio

In a geometric sequence, every term after the first is found by multiplying the previous term by a constant value, known as the common ratio, denoted by \( r \).

• This ratio is crucial because it determines the behavior and pattern of the sequence.
• The same multiplicative factor, \( r \), applies uniformly between any two consecutive terms.

For our exercise, the sequence is \( a_n = \frac{1}{4^n} \). We identified the first five terms: \( \frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \frac{1}{256}, \frac{1}{1024} \). By calculating the ratio, \( r = \frac{a_{n+1}}{a_n} \), between terms, you observe a consistent factor:

• \( \frac{1}{16} \div \frac{1}{4} = \frac{1}{4} \)
• \( \frac{1}{64} \div \frac{1}{16} = \frac{1}{4} \)
• \( \frac{1}{256} \div \frac{1}{64} = \frac{1}{4} \)
• \( \frac{1}{1024} \div \frac{1}{256} = \frac{1}{4} \)

Thus, the common ratio for this sequence is \( \frac{1}{4} \). This constant value confirms the sequence is geometric and helps to define its progression.

Nth Term Formula

The \(n\)th term formula for a geometric sequence helps us calculate any term in the sequence without listing all previous terms.

• The formula is \( a_n = a_1 \cdot r^{n-1} \).
• Here, \(a_1\) is the first term and \(r\) is the common ratio.

For the given problem, where the first term \(a_1 = \frac{1}{4}\) and the common ratio \(r = \frac{1}{4}\), the formula becomes:\[a_n = \frac{1}{4} \left( \frac{1}{4} \right)^{n-1}\]This formula provides a simple way to find any term in the sequence: just plug in the value of \(n\) you want.

• For instance, \(a_5\) would be: \( \frac{1}{4} \left( \frac{1}{4} \right)^{5-1} = \frac{1}{1024} \).

This form allows computing terms quickly and seeing how the sequence decreases rapidly with each increase in \( n \).

Sequences and Series

Understanding sequences and series is a fundamental part of mathematics, particularly in algebra and calculus. A sequence is an ordered list of numbers following a particular pattern. When we talk about geometric sequences, each term is the product of the previous term multiplied by the common ratio.

• In the given sequence, the pattern is exponential decay because each term is a smaller fraction of \( a_1 \) due to the common ratio \( \frac{1}{4} \).
• This geometric sequence will continue indefinitely if we keep increasing \( n \).

A series, on the other hand, is the sum of the terms of a sequence.

• Though our example focuses on a sequence, methods for summing geometric series exist as well.
• For geometric series with \(|r| < 1\), the series converges and has a sum, represented by \( S = \frac{a_1}{1 - r} \).
• However, our task primarily investigates the sequence pattern, not the series sum.

This understanding opens doors to exploring deeper mathematical concepts and applications.