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Chapter 15: Problem 17

Given the general term of each sequence, find each of the following. \(a_{n}=\frac{n-4}{n+6}\) a) \(a_{1}\) b) \(a_{2}\) c) the 16 th term

Short Answer

Expert verified
The answers are: a) \(a_{1} = \frac{-3}{7}\) b) \(a_{2} = \frac{-1}{4}\) c) \(a_{16} = \frac{6}{11}\)

Step by step solution

01

Find \(a_{1}\)

We're given the general term \(a_{n} = \frac{n-4}{n+6}\). To find the first term \(a_{1}\), we need to substitute \(n=1\) into the formula: \(a_{1} = \frac{1-4}{1+6}\)
02

Calculate \(a_{1}\)

Now, perform the arithmetic operations in the numerator and the denominator, and then simplify the fraction: \(a_{1} = \frac{-3}{7}\)
03

Find \(a_{2}\)

To find the second term \(a_{2}\), we need to substitute \(n=2\) into the formula: \(a_{2} = \frac{2-4}{2+6}\)
04

Calculate \(a_{2}\)

Perform the arithmetic operations in the numerator and the denominator, and then simplify the fraction: \(a_{2} = \frac{-2}{8}\) \(a_{2} = \frac{-1}{4}\)
05

Find the 16th term

To find the 16th term, we need to substitute \(n=16\) into the formula: \(a_{16} = \frac{16-4}{16+6}\)
06

Calculate the 16th term

Perform the arithmetic operations in the numerator and the denominator, and then simplify the fraction: \(a_{16} = \frac{12}{22}\) \(a_{16} = \frac{6}{11}\) The answers are: a) \(a_{1} = \frac{-3}{7}\) b) \(a_{2} = \frac{-1}{4}\) c) \(a_{16} = \frac{6}{11}\)

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