Answer:
[tex]\textsf{A)}\quad \dfrac{5}{10}, \:\: -\dfrac{6}{11}, \:\: \dfrac{7}{12}, \:\: -\dfrac{8}{13}[/tex]
[tex]\textsf{B)} \quad \text{f}\:\!(n)=\dfrac{n}{n+5}[/tex]
If n is odd then f(n) is positive.
If n is even then f(n) is negative.
C) positive
Step-by-step explanation:
Given sequence:
[tex]\dfrac{1}{6},\:\: -\dfrac{2}{7},\:\: \dfrac{3}{8}, \:\:-\dfrac{4}{9},...[/tex]
Part A
From inspection of the sequence the pattern is:
- Add 1 to the numerator of the previous term.
- Add 1 to the denominator of the previous term.
- Make every other term negative.
Therefore, assuming the pattern continues, the next four terms in the sequence will be:
[tex]\dfrac{5}{10}, \:\: -\dfrac{6}{11}, \:\: \dfrac{7}{12}, \:\: -\dfrac{8}{13}[/tex]
Part B
An explicit formula for an arithmetic sequence allows you to find the nth term of the sequence.
[tex]\textsf{First term}: \quad \text{f}(1)=\dfrac{1}{6}[/tex]
[tex]\textsf{Second term}: \quad \text{f}(2)=-\dfrac{2}{7}[/tex]
From inspection of the sequence, we can see that the numerator is the position of the term (n) in the sequence. So when n = 1 the numerator is 1, when n = 2 the numerator is 2, etc.
The denominator is 5 more than the position of the term in the sequence. So when n = 1 the denominator is 1 + 5 = 6, and when n = 2 the denominator is 2 + 5 = 7.
We can also observe that if the numerator (the value of n) is odd then the term is positive, and if the numerator (the value of n) is even then the term is negative.
Therefore, the rule for the nth term is:
[tex]\text{f}\:\!(n)=\dfrac{n}{n+5}[/tex]
If n is odd then f(n) is positive.
If n is even then f(n) is negative.
Part C
53 is an odd number.
As the function f(n) is positive when n is odd, the sign of f(53) is positive.
