Certainly! Let's solve this problem step-by-step using the properties of arithmetic sequences.
Given:
1. The second term of the sequence: [tex]\(x + 1\)[/tex]
2. The sixth term of the sequence: [tex]\(-x + 17\)[/tex]
3. The common difference of the sequence: 5
First, let's recall the formula for the [tex]\(n\)[/tex]-th term of an arithmetic sequence:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
where [tex]\(a_1\)[/tex] is the first term and [tex]\(d\)[/tex] is the common difference.
### Step 1: Express the given terms using the arithmetic sequence formula
For the second term ([tex]\(a_2\)[/tex]):
[tex]\[ a_2 = a_1 + (2-1)d \][/tex]
[tex]\[ a_2 = a_1 + d \][/tex]
We know that [tex]\(a_2 = x + 1\)[/tex],
so:
[tex]\[ x + 1 = a_1 + d \quad \text{(Equation 1)} \][/tex]
For the sixth term ([tex]\(a_6\)[/tex]):
[tex]\[ a_6 = a_1 + (6-1)d \][/tex]
[tex]\[ a_6 = a_1 + 5d \][/tex]
We know that [tex]\(a_6 = -x + 17\)[/tex],
so:
[tex]\[ -x + 17 = a_1 + 5d \quad \text{(Equation 2)} \][/tex]
### Step 2: Substitute the given common difference [tex]\(d = 5\)[/tex] into the equations
From Equation 1:
[tex]\[ x + 1 = a_1 + 5 \][/tex]
[tex]\[ a_1 = x + 1 - 5 \][/tex]
[tex]\[ a_1 = x - 4 \quad \text{(Equation 3)} \][/tex]
From Equation 2:
[tex]\[ -x + 17 = a_1 + 5 \times 5 \][/tex]
[tex]\[ -x + 17 = a_1 + 25 \][/tex]
Substitute [tex]\(a_1\)[/tex] from Equation 3:
[tex]\[ -x + 17 = (x - 4) + 25 \][/tex]
[tex]\[ -x + 17 = x - 4 + 25 \][/tex]
[tex]\[ -x + 17 = x + 21 \][/tex]
### Step 3: Solve the equation for [tex]\(x\)[/tex]
Combine like terms:
[tex]\[ -x + 17 = x + 21 \][/tex]
Bring all [tex]\(x\)[/tex] terms to one side:
[tex]\[ 17 - 21 = x + x \][/tex]
[tex]\[ -4 = 2x \][/tex]
[tex]\[ x = -2 \][/tex]
So, the value of [tex]\(x\)[/tex] is [tex]\(-2\)[/tex].
Thus, the value of [tex]\(x\)[/tex] is [tex]\(-2\)[/tex].
