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Answer:
we conclude that the rule will be:
[tex]a_n=\frac{31}{12}+\frac{1}{6}n[/tex]
Step-by-step explanation:
Given
[tex]a_6=3\frac{7}{12}=\frac{43}{12}[/tex]
[tex]a_1=2\frac{3}{4}=\frac{11}{4}[/tex]
We know the arithmetic sequence with the common difference is defined as
[tex]a_n=a_1+\left(n-1\right)d[/tex]
where a₁ is the first term and d is a common difference.
so
a₆ = a₁ + (6-1) d
substituting a₆ = 43/12 and a₁ = 11/4 to determine d
[tex]\frac{43}{12}=\:\frac{11}{4}\:+\:5d[/tex]
switch sides
[tex]\frac{11}{4}+5d=\frac{43}{12}[/tex]
subtract 11/4 from both sides
[tex]\frac{11}{4}+5d-\frac{11}{4}=\frac{43}{12}-\frac{11}{4}[/tex]
[tex]5d=\frac{5}{6}[/tex]
Divide both sides by 5
[tex]\frac{5d}{5}=\frac{\frac{5}{6}}{5}[/tex]
[tex]d=\frac{1}{6}[/tex]
as
a₁ = 11/4
[tex]d=\frac{1}{6}[/tex]
Therefore, the nth term of the Arithmetic sequence will be:
[tex]a_n=a_1+\left(n-1\right)d[/tex]
substituting d = 1/6 and a₁ = 11/4
[tex]a_n=\frac{11}{4}+\left(n-1\right)\frac{1}{6}[/tex]
[tex]=\frac{11}{4}+\frac{1}{6}n-\frac{1}{6}[/tex]
[tex]=\frac{31}{12}+\frac{1}{6}n[/tex]
Therefore, we conclude that the rule will be:
[tex]a_n=\frac{31}{12}+\frac{1}{6}n[/tex]