Answer:
We have established the following sequence
6, 12, 18,...
The given is the Arithmetic sequence with the nth term
[tex]a_n=6n[/tex]
d = 6 is a common difference.
Step-by-step explanation:
Given the formula
Sₙ = 3n² + 3n
Put n = 1 to determine the sum of the first term
S₁ = 3(1)² + 3(1)
S₁ = 3 + 3
S₁ = 6
As the sum of the first term includes the only first term, so
a₁ = 6
Put n = 2 to determine the sum of the two terms
Sₙ = 3n² + 3n
S₂ = 3n² + 3n
S₂ = 3(2)² + 3(2)
S₂ = 12 + 6
S₂ = 18
as
S₂ = a₁ + a₂
substituting S₂ = 18, a₁ = 6
18 = 6 + a₂
12 = a₂
so
a₂ = 12
Put n = 3 to determine the sum of the two terms
Sₙ = 3n² + 3n
S₃ = 3(3)² + 3(3)
S₃ = 27 + 9
S₃ =36
as
S₃ = a₁ + a₂ + a₃
substitutingS₃ =36, a₁ = 6, a₂ = 12
36 = 6+12+a₃
a₃ = 36-6-12
a₃ = 18
Thus,
a₁ = 6
a₂ = 12
a₃ = 18
Therefore, the sequence becomes
a₁, a₂, a₃,...
6, 12, 18,...
Let us check the common difference 'd' between the adjacent terms
d = 12 - 6 = 6
d = 18 - 12 = 6
As the common difference 'd' between adjacent terms is the same, therefore the sequence is Arithmetic.
Thus,
Furthermore, the nth term of the Arithmetic sequence can be defined as
[tex]a_n=a_1+\left(n-1\right)d[/tex]
substituting d = 6 and a₁ = 6
[tex]a_n=6\left(n-1\right)+6[/tex]
[tex]a_n=6n[/tex]
