Question

First term: 2 3/4 sixth term: 3 7/12 what is the rule

Answer 1

Answer:

we conclude that the rule will be:

$a_n=\frac{31}{12}+\frac{1}{6}n$

Step-by-step explanation:

Given

$a_6=3\frac{7}{12}=\frac{43}{12}$

$a_1=2\frac{3}{4}=\frac{11}{4}$

We know the arithmetic sequence with the common difference is defined as

$a_n=a_1+\left(n-1\right)d$

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

$\frac{43}{12}=\:\frac{11}{4}\:+\:5d$

switch sides

$\frac{11}{4}+5d=\frac{43}{12}$

subtract 11/4 from both sides

$\frac{11}{4}+5d-\frac{11}{4}=\frac{43}{12}-\frac{11}{4}$

$5d=\frac{5}{6}$

Divide both sides by 5

$\frac{5d}{5}=\frac{\frac{5}{6}}{5}$

$d=\frac{1}{6}$

as

a₁ = 11/4

$d=\frac{1}{6}$

Therefore, the nth term of the Arithmetic sequence will be:

$a_n=a_1+\left(n-1\right)d$

substituting d = 1/6 and a₁ = 11/4

$a_n=\frac{11}{4}+\left(n-1\right)\frac{1}{6}$

    $=\frac{11}{4}+\frac{1}{6}n-\frac{1}{6}$

    $=\frac{31}{12}+\frac{1}{6}n$

Therefore, we conclude that the rule will be:

$a_n=\frac{31}{12}+\frac{1}{6}n$