Question

First term: 2 3/4 sixth term: 3 7/12

Answer 1

Answer:

We conclude that the rule will be:

aₙ = 31/12 +  1/6n

Step-by-step explanation:

Answer:

we conclude that the rule will be:

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

Step-by-step explanation:

Given

• $a_6=3\frac{7}{12}=\frac{43}{12}$

• $a_1=2\frac{3}{4}=\frac{11}{4}$

The nth term of Arithmetic Sequence

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.

To Determine:

The Rule of the nth term of Arithmetic Sequence

Steps to solve the problem

The 6th term of the Arithmetic sequence be defined as

a₆ = a₁ + (6-1) d

substituting a₆ = 43/12 and a₁ = 11/4 to determine d

43/12 = 11/4 + 5d

switch sides

11/4 + 5d = 43/12

subtract 11/4 from both sides

11/4 + 5d - 11/4 = 43/12 - 11/4

5d  = 5/6

Divide both sides by 5

5d/5 = [5/6] / [5]

d = 1/6

as

a₁ = 11/4

d = 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

aₙ = 11/4 + (n-1) × 1/6

    = 11/4 + 1/6n - 1/6

    = 31/12 +  1/6n

Therefore, we conclude that the rule will be:

aₙ = 31/12 +  1/6n