Question

Are the sequences geometric? For those that are, give a formula for the \(n^{\text {th }}\) term. $$ 1, \frac{1}{1.5}, \frac{1}{(1.5)^{2}}, \frac{1}{(1.5)^{3}}, \cdots $$

Short answer

If so, what is the formula for the nth term of the sequence? Answer: Yes, the sequence is geometric with a common ratio of $\frac{1}{1.5}$. The formula for the nth term of the sequence is $a_n = \frac{1}{(1.5)^{n-1}}$.

Step-by-step solution

Identify the ratio between consecutive terms

To find out if the sequence is geometric, we need to find the ratio between consecutive terms, i.e. \(\frac{a_{n}}{a_{n-1}}\), and check if it is constant. In this case, we have: $$ \frac{\frac{1}{(1.5)^1}}{1} = \frac{1}{1.5} \\ \frac{\frac{1}{(1.5)^2}}{\frac{1}{(1.5)^1}} = \frac{1}{1.5} \\ \frac{\frac{1}{(1.5)^3}}{\frac{1}{(1.5)^2}} = \frac{1}{1.5} \\ $$ Since the ratio between consecutive terms is constant, this sequence is indeed geometric.

Determine the first term and the common ratio

In a geometric sequence, the first term is denoted as \(a_1\), and the common ratio is often denoted as \(r\). In this case, the first term \(a_1= 1\), and the common ratio \(r = \frac{1}{1.5}\).

Find the formula for the \(n^{\text{th}}\) term

For a geometric sequence, the formula for the \(n^{\text{th}}\) term \(a_n\) is given by: $$ a_n = a_1 \cdot r^{n-1} $$ Substitute \(a_1 = 1\) and \(r = \frac{1}{1.5}\) into the formula: $$ a_n = 1 \cdot \left(\frac{1}{1.5}\right)^{n-1} $$ Or $$ a_n = \frac{1}{(1.5)^{n-1}} $$ So, the formula for the \(n^{\text{th}}\) term of the given geometric sequence is \(a_n = \frac{1}{(1.5)^{n-1}}\).

Key concepts

Common Ratio

In a geometric sequence, the common ratio is a crucial element that defines the relationship between consecutive terms. It is denoted by the letter \( r \). The common ratio is the factor by which you multiply any term in the sequence to get the next term. To determine if a sequence is geometric, calculate the ratio between consecutive terms and see if it remains constant across the entire sequence.
For instance, let's take this sequence:

• First term: \( 1 \)
• Second term: \( \frac{1}{1.5} \)
• Third term: \( \frac{1}{(1.5)^2} \)

The common ratio \( r \) can be calculated as \( \frac{\text{second term}}{\text{first term}} = \frac{1}{1.5} \).
If this ratio is constant for all consecutive terms, the sequence is geometric.
In our example, the ratio \( \frac{1}{1.5} \) persists consistently, confirming the sequence is indeed geometric.

Formula for nth Term

One of the most powerful tools for working with geometric sequences is the formula for the \( n^{\text{th}} \) term. This formula lets you find any term in a geometric sequence without listing out all the previous terms. The formula is written as:

• \( a_n = a_1 \cdot r^{n-1} \)

Here:

• \( a_n \) is the \( n^{\text{th}} \) term.
• \( a_1 \) is the first term.
• \( r \) is the common ratio.
• \( n \) is the term number.

For the sequence given, where \( a_1 = 1 \) and \( r = \frac{1}{1.5} \), the \( n^{\text{th}} \) term formula becomes: \( a_n = 1 \cdot \left(\frac{1}{1.5}\right)^{n-1} \).
This means that for any position \( n \), you can quickly calculate the term using this expression.

Geometric Progression

A geometric progression is a sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. Geometric sequences are everywhere, from the expanding universe to calculating compound interest. Recognizing them is invaluable for both theoretical and practical applications.
Here's what makes something a geometric progression:

• All terms are connected through multiplication by a common ratio.
• The sequence can grow rapidly (if \( r > 1 \)) or shrink (if \( 0 < r < 1 \)).

Our example sequence \( 1, \frac{1}{1.5}, \frac{1}{(1.5)^2} \), showcases a geometric progression as it maintains a constant common ratio of \( \frac{1}{1.5} \). With this background, you can identify any geometric progression and use its properties to solve various mathematical problems.