Problem 60 - Algebra 2

Question

Find \(a_{1}\) for a geometric sequence with the given terms. $$ a_{9}=\frac{1}{2} \text { and } a_{12}=\frac{1}{16} $$

Step-by-Step Solution

Answer

\(\ a_{1} = 128\)

Step 1: Identify the common ratio

From the problem, we know that the ninth term (\(a_{9}\)) is \(\frac{1}{2}\) and the twelfth term (\(a_{12}\)) is \(\frac{1}{16}\). The ratio between every pair of terms in a geometric sequence always remains the same. We can take advantage of this property to find the ratio, \(r\). According to this formula: \[a_{12} = a_{9} \cdot r^{12-9}\] Solve for \(r\), remembering that \(a_{12}\) is \(\frac{1}{16}\) and \(a_{9}\) is \(\frac{1}{2}\).

Step 2: Compute the ratio

Replace the values of \(a_{12}\) and \(a_{9}\) in the formula and solve for \(r\).\[\frac{1}{16} = \frac{1}{2} * r^{12-9}\]\[r^{3} = \frac{1}{32}\]\[r = \sqrt[3]{\frac{1}{32}} = \frac{1}{2}\]

Step 3: Determine the value of \(a_{1}\)

Now that we've found \(r\), we can use it to find \(a_{1}\). The nth term of a geometric sequence is given by \(a_{n} = a_{1} \cdot r^{n-1}\). Let's isolate \(a_{1}\) and replace \(a_{9}\), \(r\), and \(n=9\). Since we know that\(a_{9} = \frac{1}{2}\), \(r = \frac{1}{2}\) and \(n = 9\) in the equation \[a_{1} = a_{9} / r^{9-1}\]

Step 4: Solve to find \(a_{1}\)

Replace the known values in the previous equation\[\frac{1}{2} = a_{1} * (\frac{1}{2})^{9-1}\]\[a_{1} = \frac{1}{2} / (\frac{1}{2})^{8}\]\[a_{1} = \frac{1}{2} * \frac{2^{8}}{1}\]\[a_{1} = 128\]

Key Concepts

Common Ratio

In a geometric sequence, the common ratio is a crucial element. It is the factor by which you multiply one term to get the next. This value, denoted as \(r\), remains consistent throughout the sequence. In our exercise, we find the common ratio by using two terms: \(a_{9} = \frac{1}{2}\) and \(a_{12} = \frac{1}{16}\). By applying the formula \(a_{n} = a_{m} \cdot r^{n-m}\), we solve for \(r\).

In this instance, substituting the known values yields \(r^3 = \frac{1}{32}\) and further solving gives \(r = \frac{1}{2}\). Knowing how to calculate the common ratio helps you understand how each term relates to its neighbors.

Nth Term

Learning to identify the nth term \((a_{n})\) is vital when working with geometric sequences. The nth term is any specific position within a sequence. Generally, it is expressed with the formula:

• \(a_{n} = a_{1} \cdot r^{n-1}\)

The nth term allows you to find the value at any position without listing all previous terms.

In the given problem, we are tasked with finding \(a_1\) using \(a_{9}\) and \(r\). It involves backtracking through the sequence to determine the first term. This kind of calculation often requires manipulating the original nth term formula to solve for different variables.

Geometric Progression

A geometric progression is a sequence where each term is derived by multiplying the previous one by the common ratio. This exponential nature contrasts with arithmetic sequences, where the same constant is added each time. Geometric progression is fundamental in understanding growth patterns in nature and finance.

Working through this exercise exemplifies key aspects of geometric progression: calculating terms with specific knowns and determining early elements from later ones. The consistency of the common ratio ensures predictable, repeatable patterns throughout the sequence, making it easier to comprehend and calculate unknown terms.

Sequence Calculation

Calculating a sequence involves using the known attributes like the first term \((a_1)\), common ratio \((r)\), and any known terms \((a_n)\). The exercise illustrates that sequence calculation comprises several steps:

• Identifying the relationship between terms.
• Using that relationship to define an equation representing the sequence.
• Solving for unknowns with the help of algebraic manipulation.

Such calculations are not just about math but also mastering strategic approaches and processes. Ultimately, this translates into solving larger problems, whether academic or in real-world applications, using a clear logical path and mathematical principles.