Question

Show that each sequence is geometric. Then find the common ratio and list the first four terms. $$ \left\\{t_{n}\right\\}=\left\\{\frac{3^{n-1}}{2^{n}}\right\\} $$

Short answer

The common ratio is \(\frac{3}{2}\). The first four terms are \(\frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}\).

Step-by-step solution

Identify the General Term

The general term of the sequence is given by: \(t_n = \frac{3^{n-1}}{2^n}\)

Find the First Four Terms

Substitute the values of \(n = 1, 2, 3, 4\) to find the first four terms:For \(n = 1\), \(t_1 = \frac{3^{1-1}}{2^1} = \frac{3^0}{2^1} = \frac{1}{2}\)For \(n = 2\), \(t_2 = \frac{3^{2-1}}{2^2} = \frac{3^1}{2^2} = \frac{3}{4}\)For \(n = 3\), \(t_3 = \frac{3^{3-1}}{2^3} = \frac{3^2}{2^3} = \frac{9}{8}\)For \(n = 4\), \(t_4 = \frac{3^{4-1}}{2^4} = \frac{3^3}{2^4} = \frac{27}{16}\)

Confirm It is Geometric

To confirm the sequence is geometric, the ratio between consecutive terms must be constant:\(r = \frac{t_{n+1}}{t_n}\)Compute for first couple of terms:\(r = \frac{t_2}{t_1} = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{2}\)\(r = \frac{t_3}{t_2} = \frac{\frac{9}{8}}{\frac{3}{4}} = \frac{3}{2}\)\(r = \frac{t_4}{t_3} = \frac{\frac{27}{16}}{\frac{9}{8}} = \frac{3}{2}\)Since the ratio \(r\) is constant and equal to \(\frac{3}{2}\), the sequence is geometric.

State the Common Ratio and First Four Terms

The common ratio \(r\) is \(\frac{3}{2}\). The first four terms of the sequence are \(\frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}\).

Key concepts

Common Ratio

In a geometric sequence, the common ratio is the constant factor between consecutive terms. If you have a sequence where each term after the first is found by multiplying the previous term by the same number, you have a geometric sequence.
In this exercise, the common ratio is found by dividing one term by the previous term, showing us how much each term is scaled from the last.
For example, to find the common ratio of the given sequence, we divide the second term by the first term:
\( r = \frac{t_2}{t_1}\). Substituting the given values:
\( t_2 = \frac{3}{4} \) and \( t_1 = \frac{1}{2} \), we get:
\[ r = \frac{\frac{3}{4}}{\frac{1}{2}} = \frac{3}{2} \]
By confirming this ratio is consistent for subsequent terms, we see that the common ratio for our sequence is indeed \( \frac{3}{2} \).

General Term

The general term of a sequence describes a formula that can generate any term in the sequence based on its position number. Understanding the general term allows us to find any term in the sequence without listing all previous terms.
In our exercise, the general term of the sequence is given by:
\( t_n = \frac{3^{n-1}}{2^n} \)
This formula lets us calculate any term by substituting 'n' with the position number. For instance:
- When \( n = 1 \), substituting into the general term gives us:
\[ t_1 = \frac{3^{1-1}}{2^1} = \frac{3^0}{2^1} = \frac{1}{2} \]
This formula is powerful as it directly correlates the term's position with its value.

Sequence Terms

Sequence terms are the individual elements that make up the sequence. In our exercise:
- The first term \( t_1 \) is \( \frac{1}{2} \).
- The second term \( t_2 \) is \( \frac{3}{4} \).
- The third term \( t_3 \) is \( \frac{9}{8} \).
- The fourth term \( t_4 \) is \( \frac{27}{16} \).
Using the general term \( t_n = \frac{3^{n-1}}{2^n} \), each term is calculated by substituting the respective position number 'n'. The sequence starts with \( t_1 = \frac{1}{2} \) and each subsequent term grows by a factor of \( \frac{3}{2} \).

Algebra in Geometric Sequences

Algebra plays a crucial role in understanding geometric sequences. By using algebraic formulas and properties, we can define the behavior of the sequence.
The formula \( t_n = \frac{3^{n-1}}{2^n} \) is an example of algebra at work, showing the relationship between the term's position and its value. Proper algebraic manipulation helps us determine the common ratio and verify the geometric nature of the sequence by consistently showing the same ratio between terms.
For example, confirming the common ratio through algebraic steps:
\[ r = \frac{\frac{9}{8}}{\frac{3}{4}} = \frac{9/8}{3/4} = \frac{9}{8} \times \frac{4}{3} = \frac{36}{24} = \frac{3}{2} \]
These algebraic calculations validate that each term grows by a consistent factor, proving the sequence's geometric nature.