Question

1. the ( n^{\text{th}} ) term of a sequence is given by ( a_n = 7 - 2(n - 6) ).

a. is the sequence arithmetic, geometric, or neither? explain.
it is an arithmetic because it has a constant ratio between terms.
b. find ( a_6 ).

Explanation:

Step1: Simplify the formula for \(a_n\)

First, expand the given formula \(a_n = 7 - 2(n - 6)\). Using the distributive property \(a(b - c)=ab - ac\), we get:
\(a_n=7-2n + 12\)
Combine like terms: \(a_n=-2n + 19\)

Step2: Analyze part (a)

For a sequence to be arithmetic, the difference between consecutive terms (\(a_{n + 1}-a_n\)) should be constant. Let's find \(a_{n+1}\):
\(a_{n + 1}=-2(n + 1)+19=-2n-2 + 19=-2n + 17\)
Now, find \(a_{n + 1}-a_n\):
\((-2n + 17)-(-2n + 19)=-2n + 17 + 2n-19=-2\)
Since the common difference \(d=-2\) is constant, the sequence is arithmetic. (Note: The original explanation had a mistake saying "constant ratio" - arithmetic sequences have constant difference, geometric have constant ratio)

Step3: Solve part (b) to find \(a_6\)

Substitute \(n = 6\) into the formula \(a_n=-2n + 19\) (or the original formula \(a_n = 7-2(n - 6)\)):
Using \(a_n = 7-2(n - 6)\) with \(n = 6\):
\(a_6=7-2(6 - 6)=7-2(0)=7\)
Or using \(a_n=-2n + 19\) with \(n = 6\):
\(a_6=-2(6)+19=-12 + 19 = 7\)

Answer:

a. The sequence is arithmetic because the difference between consecutive terms (\(a_{n + 1}-a_n\)) is constant (\(-2\)).
b. \(a_6 = 7\)