Problem 20 - Algebra for College Students

Question

\text { The } 9 \text { th term of } \frac{16}{81}, \frac{8}{27}, \frac{4}{9}, \frac{2}{3}, \ldots . \frac{81}{16}

Step-by-Step Solution

Answer

The 9th term is \( \frac{16}{531441} \).

Step 1: Identify the Sequence Type

The given sequence is \( \frac{16}{81}, \frac{8}{27}, \frac{4}{9}, \frac{2}{3}, \ldots \). Notice that each term is a fraction and is decreasing in a specific way. This indicates it's a geometric sequence where each term is obtained by multiplying the previous term by a constant ratio.

Step 2: Find the Common Ratio

To find the common ratio \( r \), divide the second term by the first: \( r = \frac{8/27}{16/81} = \frac{8}{27} \times \frac{81}{16} = \frac{1}{3} \). Thus, the common ratio of the sequence is \( \frac{1}{3} \).

Step 3: Write the General Formula for a Geometric Sequence

The general term (\( a_n \)) of a geometric sequence is given by: \( a_n = a_1 \times r^{(n-1)} \), where \( a_1 \) is the first term and \( r \) is the common ratio.

Step 4: Substitute Known Values

Here, the first term \( a_1 = \frac{16}{81} \), and the common ratio \( r = \frac{1}{3} \). Substitute these into the formula to get \( a_n = \frac{16}{81} \times \left(\frac{1}{3}\right)^{(n-1)} \).

Step 5: Calculate the 9th Term

Substitute \( n = 9 \) into the formula: \( a_9 = \frac{16}{81} \times \left(\frac{1}{3}\right)^{8} = \frac{16}{81} \times \frac{1}{6561} \). This simplifies to \( \frac{16}{531441} \).

Step 6: Simplify the Expression

Simplify \( \frac{16}{531441} \) by evaluating if it can be reduced further. Here, both the numerator and the denominator don't share common factors other than 1, so it is already in its simplest form.

Key Concepts

Common Ratio

In a geometric sequence, understanding the concept of a common ratio is key. A geometric sequence is defined where each term after the initial one is found by multiplying the previous term by a constant. This constant is what we call the common ratio.
Take our sequence:

• First term (\(a_1\) is \(\frac{16}{81}\)).
• Second term (\(a_2\) is \(\frac{8}{27}\)).

To find the common ratio \(r\), we divide the second term by the first term:
\[ r = \frac{8/27}{16/81} = \frac{8}{27} \times \frac{81}{16} = \frac{1}{3}.\]This means, in our sequence, each term is obtained by multiplying the previous one by \(\frac{1}{3}\). This factor of \(\frac{1}{3}\) is what keeps the sequence uniformly shrinking.

General Term

The beauty of a geometric sequence lies in its general term formula. Knowing this helps you find any term in the sequence without listing them all. The general term \(a_n\) of a geometric sequence can be expressed as:
\[a_n = a_1 \times r^{(n-1)},\]where \(a_1\) is the first term, and \(r\) is the common ratio.
For our sequence, the first term is \(\frac{16}{81}\), and we've determined the common ratio to be \(\frac{1}{3}\). Plugging these into the formula gives:

• \(a_n = \frac{16}{81} \times \left(\frac{1}{3}\right)^{(n-1)}\).

This general term equation allows for easy computation of any term within the sequence simply by changing \(n\).

Nth Term

The nth term formula in a geometric sequence exemplifies the power of mathematics in simplifying patterns. When asked to find a specific term, such as the 9th term in our sequence, the general term formula comes into play:
\[a_n = \frac{16}{81} \times \left(\frac{1}{3}\right)^{(n-1)}.\]To find the 9th term, we substitute \(n = 9\) into the formula:

• \(a_9 = \frac{16}{81} \times \left(\frac{1}{3}\right)^{8}\).

Calculate further to get:
\[a_9 = \frac{16}{81} \times \frac{1}{6561} = \frac{16}{531441}.\]Upon simplification, \(\frac{16}{531441}\) results in a neat, simplified form, marking the 9th term of our sequence.