Problem 17 - Beginning and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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: \(a_{1} = \frac{1-4}{1+6}\)

Step 2: Calculate \(a_{1}\)

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

Step 3: Find \(a_{2}\)

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

Step 4: Calculate \(a_{2}\)

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

Step 5: Find the 16th term

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

Step 6: 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) \(a_{16} = \frac{6}{11}\)