Problem 35 - Precalculus: Mathematical for Calculus

Question

The common ratio in a geometric sequence is \(\frac{2}{5},\) and the fourth term is \(\frac{5}{2} .\) Find the third term.

Step-by-Step Solution

Answer

The third term is \(\frac{25}{4}\).

Step 1: Understanding the Geometric Sequence

A geometric sequence is defined by a first term, \(a_1\), and a common ratio, \(r\). Each term, \(a_n\), in the sequence is found by multiplying the previous term by the common ratio. Thus, the formula for the nth term in a geometric sequence is given by \(a_n = a_1 \cdot r^{n-1}\).

Step 2: Identify Given Information

We know that the common ratio \(r\) is \(\frac{2}{5}\) and the fourth term \(a_4\) is \(\frac{5}{2}\). We need to find the third term \(a_3\).

Step 3: Write the Formula for the Fourth Term

Using the formula for the nth term, write the expression for the fourth term: \[ a_4 = a_1 \cdot r^3 \]. Given \(a_4 = \frac{5}{2}\), substitute the known values into the equation: \[ \frac{5}{2} = a_1 \cdot \left( \frac{2}{5} \right)^3 \].

Step 4: Solve for the First Term

Calculate \(\left( \frac{2}{5} \right)^3\): \[ \left( \frac{2}{5} \right)^3 = \frac{8}{125} \]. Then solve \( \frac{5}{2} = a_1 \cdot \frac{8}{125} \) for \(a_1\) by multiplying both sides by \(\frac{125}{8}\) to isolate \(a_1\): \[ a_1 = \frac{5}{2} \cdot \frac{125}{8} = \frac{625}{16} \].

Step 5: Calculate the Third Term

Now use the formula for the third term: \( a_3 = a_1 \cdot r^2 \). Substitute \(a_1 = \frac{625}{16}\) and \(r = \frac{2}{5}\): \[ a_3 = \frac{625}{16} \cdot \left( \frac{2}{5} \right)^2 = \frac{625}{16} \cdot \frac{4}{25} \].

Step 6: Simplify to Find the Third Term

Simplify \( \frac{625}{16} \cdot \frac{4}{25}\):\[ \frac{625}{16} \cdot \frac{4}{25} = \frac{625 \cdot 4}{16 \cdot 25} = \frac{2500}{400} = \frac{25}{4} \]. Thus, the third term is \(\frac{25}{4}\).

Key Concepts

Common Ratio

In a geometric sequence, the common ratio is a fundamental concept that sets each term in relation to its predecessor. It's the constant factor that you multiply by to get from one term to the next. This ratio is usually denoted by the symbol \( r \). For example, if the first term of a sequence is 6 and the common ratio is 2, the sequence will be 6, 12, 24, 48, and so on.

Understanding the common ratio helps not only in recognizing a geometric sequence but also in predicting its pattern. Here’s how you can identify and use the common ratio:

• **Consistent Factor:** Every term after the first is produced by multiplying the previous term by \( r \).
• **Backward Calculation:** You can also find a previous term by dividing a term by \( r \).
• **Critical for Formulas:** It appears in the formula for the nth term and influences the behavior of the sequence significantly.

In the exercise example, the common ratio is \( \frac{2}{5} \), meaning each term is \( \frac{2}{5} \) times the term before it.

Nth Term Formula

The nth term formula is a key expression in understanding geometric sequences. It allows you to find any term in the sequence without listing out all the previous terms. The formula is written as:

\[ a_n = a_1 \cdot r^{n-1} \]

Here's a breakdown:

• **\( a_n \):** The nth term you're searching for.
• **\( a_1 \):** The first term of the sequence.
• **\( r \):** The common ratio of the sequence.
• **\( n \):** The position of the term in the sequence.

By using this formula, you can jump straight to the term you need. For instance, if you know the first term \( a_1 \) is 8 and the common ratio \( r \) is 3, you can find the fourth term by using

\[ a_4 = 8 \cdot 3^{4-1} = 8 \cdot 27 = 216 \].

In our exercise problem, this formula helped find the third term by plugging in the known values and solving algebraically.

Arithmetic Manipulation

Arithmetic manipulation involves using basic arithmetic operations to solve for unknowns in equations, especially in sequences. This skill is crucial when working with formulas to simplify and solve them properly.

Here are some arithmetic manipulation examples used in the exercise:

• **Exponentiation:** Calculating \( \left( \frac{2}{5} \right)^3 \) involves multiplying \( \frac{2}{5} \) by itself twice, resulting in \( \frac{8}{125} \).
• **Isolating Variables:** To find \( a_1 \) from \( \frac{5}{2} = a_1 \cdot \frac{8}{125} \), multiply both sides by \( \frac{125}{8} \) to isolate \( a_1 \).
• **Simplification:** When calculating \( a_3 \), once you've reached an expression like \( \frac{625}{16} \cdot \frac{4}{25} \), multiplying the numerators and denominators gives \( \frac{2500}{400} \), which can be simplified to \( \frac{25}{4} \).

These operations are essential in manipulating equations to find the terms of interest efficiently and correctly.