Let's break this down and solve each part step-by-step.
### Given:
- The third term [tex]\((a_3)\)[/tex] = [tex]\(5x - 8\)[/tex]
- The fifth term [tex]\((a_5)\)[/tex] = [tex]\(x^2 - 2x\)[/tex]
- The seventh term [tex]\((a_7)\)[/tex] = [tex]\(3x - 10\)[/tex]
### Part (a) Find the value of [tex]\(x\)[/tex].
In an arithmetic sequence, the terms can be expressed in terms of the first term [tex]\(a_1\)[/tex] and the common difference [tex]\(d\)[/tex]:
- [tex]\(a_3 = a_1 + 2d\)[/tex]
- [tex]\(a_5 = a_1 + 4d\)[/tex]
- [tex]\(a_7 = a_1 + 6d\)[/tex]
Given:
[tex]\[
a_3 = 5x - 8
\][/tex]
[tex]\[
a_5 = x^2 - 2x
\][/tex]
[tex]\[
a_7 = 3x - 10
\][/tex]
From the sequence we know:
[tex]\[
a_5 = a_3 + 2d
\][/tex]
Substitute [tex]\(a_3\)[/tex]:
[tex]\[
x^2 - 2x = (5x - 8) + 2d
\][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[
x^2 - 2x - 5x + 8 = 2d
\][/tex]
[tex]\[
x^2 - 7x + 8 = 2d
\][/tex]
[tex]\[
d = \frac{x^2 - 7x + 8}{2}
\][/tex]
Next, from:
[tex]\[
a_7 = a_3 + 4d
\][/tex]
Substitute [tex]\(a_3\)[/tex]:
[tex]\[
3x - 10 = (5x - 8) + 4d
\][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[
3x - 10 - 5x + 8 = 4d
\][/tex]
[tex]\[
-2x - 2 = 4d
\][/tex]
[tex]\[
d = \frac{-2x - 2}{4}
\][/tex]
[tex]\[
d = \frac{-x - 1}{2}
\][/tex]
Set the expressions for [tex]\(d\)[/tex] equal to each other:
[tex]\[
\frac{x^2 - 7x + 8}{2} = \frac{-x - 1}{2}
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
x^2 - 7x + 8 = -x - 1
\][/tex]
[tex]\[
x^2 - 6x + 9 = 0
\][/tex]
[tex]\[
(x - 3)^2 = 0
\][/tex]
[tex]\[
x = 3
\][/tex]
### Part (b) Find the common difference [tex]\(d\)[/tex] and the term of the series.
Substitute [tex]\(x = 3\)[/tex] back into the expression for [tex]\(d\)[/tex]:
[tex]\[
d = \frac{-3 - 1}{2} = \frac{-4}{2} = -2
\][/tex]
First term, [tex]\(a_1\)[/tex]:
[tex]\[
a_1 = a_3 - 2d
\][/tex]
[tex]\[
a_3 = 5 \cdot 3 - 8 = 15 - 8 = 7
\][/tex]
[tex]\[
a_1 = 7 - 2(-2)
\][/tex]
[tex]\[
a_1 = 7 + 4 = 11
\][/tex]
### Part (c) Calculate the sum of the first 20 terms of the series.
Sum of the first [tex]\(n\)[/tex] terms in an arithmetic series:
[tex]\[
S_n = \frac{n}{2} [2a_1 + (n-1)d]
\][/tex]
Substitute [tex]\(n = 20\)[/tex], [tex]\(a_1 = 11\)[/tex], [tex]\(d = -2\)[/tex]:
[tex]\[
S_{20} = \frac{20}{2} [2 \cdot 11 + (20-1) \cdot (-2)]
\][/tex]
[tex]\[
S_{20} = 10 [22 + 19 \cdot (-2)]
\][/tex]
[tex]\[
S_{20} = 10 [22 - 38]
\][/tex]
[tex]\[
S_{20} = 10 \cdot (-16)
\][/tex]
[tex]\[
S_{20} = -160
\][/tex]
Thus, the sum of the first 20 terms is [tex]\(-160\)[/tex].
### Final Answers:
(a) The value of [tex]\(x\)[/tex] is [tex]\(3\)[/tex].
(b) The common difference [tex]\(d\)[/tex] is [tex]\(-2\)[/tex] and the first term [tex]\(a_1\)[/tex] is [tex]\(11\)[/tex].
(c) The sum of the first 20 terms is [tex]\(-160\)[/tex].
