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Chapter 7: Problem 36

Find the tenth term of a geometric sequence whose second term is 3 and whose seventh term is 11 .

Short Answer

Expert verified
The tenth term of the geometric sequence is approximately \(a_{10} \approx 21.708\).

Step by step solution

01

Identify given terms

We are given the second term a_2 = 3 and the seventh term a_7 = 11.
02

Write down equations using the geometric sequence formula

We can now write down the equations for the second term and the seventh term using the geometric sequence formula: \(a_2 = a_1 * r^{(2-1)} = a_1 * r \) \(a_7 = a_1 * r^{(7-1)} = a_1 * r^6 \) Substitute the given values for a_2 and a_7: \(3 = a_1 * r\) \(11 = a_1 * r^6 \)
03

Solve for the common ratio r and the first term a_1

First, solve the equation for a_1 in terms of r: Divide the second equation by the first equation: \(\frac{11}{3} = \frac{a_1 * r^6}{a_1 * r} \) The \(a_1\) terms cancel out, and we are left with: \(\frac{11}{3} = r^5\) Now, solve for r: \(r = \sqrt[5]{\frac{11}{3}}\) Next, solve for a_1 using the equation for the second term: \(3 = a_1 * \sqrt[5]{\frac{11}{3}}\) Solve for a_1: \(a_1 = \frac{3}{\sqrt[5]{\frac{11}{3}}}\)
04

Find the tenth term using the geometric sequence formula

Now that we have the first term a_1 and the common ratio r, we can find the tenth term using the geometric sequence formula: \(a_{10} = a_1 * r^{(10-1)} = a_1 * r^9\) Substitute the values for a_1 and r: \(a_{10} = \frac{3}{\sqrt[5]{\frac{11}{3}}} * \left(\sqrt[5]{\frac{11}{3}}\right)^9\) Simplify the expression: \(a_{10} = 3 * \left(\frac{11}{3}\right)^{\frac{9}{5}}\) Calculate the answer: \(a_{10} \approx 21.708\) So, the tenth term of the geometric sequence is approximately 21.708.

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