Question

Write the algebraic form of this sequence.

Answer 1

We need to find the formula for the n-th term (aₙ) of the given arithmetic sequence.

The formula for the n-th term of an arithmetic sequence is aₙ = a + (n-1)d, where a is the first term and d is the common difference.

The given sequence is 5, 11, 17, ...
The first term is a = 5.
The common difference is d = 11 - 5 = 6. (We can check: 17 - 11 = 6).
Substitute these values into the formula:
aₙ = 5 + (n-1)6 Distribute the 6:
aₙ = 5 + 6n - 6 Combine the constant terms:
aₙ = 6n - 1 The algebraic form of the sequence is 6n - 1.

Answer 2

We need to find the sum of the first 15 terms (S₁₅) for the sequence 5, 11, 17, ...

The formula for the sum of the first n terms of an arithmetic sequence is:
Sₙ = (n)/(2)[2a + (n-1)d] From the previous part, we know:
First term, a = 5.
Common difference, d = 6.
Number of terms, n = 15.
Substitute these values into the sum formula:
S₁₅ = (15)/(2)[2(5) + (15-1)6] S₁₅ = (15)/(2)[10 + (14)6] S₁₅ = (15)/(2)[10 + 84] S₁₅ = (15)/(2)[94] S₁₅ = 15 × 47 S₁₅ = 705 The sum of the first 15 terms is 705.

Answer 3

We need to find the general formula for the sum of the first n terms (Sₙ) of the sequence 5, 11, 17, ...

We use the general sum formula Sₙ = (n)/(2)[2a + (n-1)d] and substitute the values of a and d.

We have a = 5 and d = 6.
Substitute these into the formula:
Sₙ = (n)/(2)[2(5) + (n-1)6] Simplify the expression inside the brackets:
Sₙ = (n)/(2)[10 + 6n - 6] Sₙ = (n)/(2)[6n + 4] Now, distribute the (n)/(2) to the terms inside the bracket:
Sₙ = n ( (6n)/(2) + (4)/(2) ) Sₙ = n(3n + 2) Sₙ = 3n² + 2n The sum of the first n terms is 3n² + 2n.