Math Problem Statement

Find an and a5 for the following geometric sequence: 9/5, 3, 5, 25/3. Simplify your answer and type an integer or a fraction.

Solution

We are tasked with solving for ana_n and a5a_5 in the given geometric sequence 95,3,5,253\frac{9}{5}, 3, 5, \frac{25}{3}. Let's proceed step by step.


Step 1: Identify the common ratio rr

To find rr, divide any term by the previous term in the geometric sequence:

  1. Between the 2nd and 1st terms: r=395=3×59=159=53.r = \frac{3}{\frac{9}{5}} = 3 \times \frac{5}{9} = \frac{15}{9} = \frac{5}{3}.

  2. Between the 3rd and 2nd terms: r = \frac{5}{3} = \frac{5}{3} \quad (\text{same ratio confirms r}).

Thus, the common ratio rr is 53\frac{5}{3}.


Step 2: Write the general term ana_n

The formula for the nn-th term of a geometric sequence is: an=a1rn1,a_n = a_1 \cdot r^{n-1}, where a1=95a_1 = \frac{9}{5} and r=53r = \frac{5}{3}. Substituting these values, we get: an=95(53)n1.a_n = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^{n-1}.


Step 3: Simplify the expression for ana_n

Simplify the general term ana_n using properties of exponents: an=95(53)n1.a_n = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^{n-1}.

  • Notice that (53)n1\left( \frac{5}{3} \right)^{n-1} stays as a power.

Thus, the formula for ana_n is: an=95(53)n1.a_n = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^{n-1}.


Step 4: Find a5a_5

To find a5a_5, substitute n=5n = 5 into ana_n:

a5=95(53)51=95(53)4.a_5 = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^{5-1} = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^4.

Simplify (53)4\left( \frac{5}{3} \right)^4: (53)4=5434=62581.\left( \frac{5}{3} \right)^4 = \frac{5^4}{3^4} = \frac{625}{81}.

Now multiply: a5=9562581.a_5 = \frac{9}{5} \cdot \frac{625}{81}. Simplify step-by-step: a5=9625581=5625405.a_5 = \frac{9 \cdot 625}{5 \cdot 81} = \frac{5625}{405}.

Simplify the fraction: a5=1259.a_5 = \frac{125}{9}.


Final Answers:

  1. General term ana_n: an=95(53)n1.a_n = \frac{9}{5} \cdot \left( \frac{5}{3} \right)^{n-1}.
  2. Fifth term a5a_5: a5=1259.a_5 = \frac{125}{9}.

Do you need further clarification or additional details? Here are 5 related questions:

  1. How do you determine the common ratio in a geometric sequence?
  2. What is the formula for the sum of a geometric sequence?
  3. How would the formula change if the common ratio were negative?
  4. Can you identify whether a sequence is geometric or arithmetic?
  5. How do you find the nn-th term if the starting term changes?

Tip: Always check for consistency by verifying the ratio between consecutive terms when dealing with geometric sequences.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Common Ratio
Exponential Functions

Formulas

General term of a geometric sequence: an = a1 * r^(n-1)

Theorems

Properties of geometric sequences

Suitable Grade Level

Grades 9-11