Answer :
Let's solve this by breaking it down into several steps.
### Part 1: Solving for the geometric series
1. Sum of the first [tex]\(n\)[/tex] terms:
We know the sum of the first [tex]\(n\)[/tex] terms of a geometric series with the first term [tex]\(a\)[/tex] and common ratio [tex]\(r\)[/tex] is:
[tex]\[ S_n = \frac{a(r^n - 1)}{r - 1} \][/tex]
Given [tex]\(S_n = 364\)[/tex] and [tex]\(a = 1\)[/tex], we have:
[tex]\[ \frac{r^n - 1}{r - 1} = 364 \quad \text{(1)} \][/tex]
2. Sum of the reciprocals of the terms:
The sum of the reciprocals of the series is given by:
[tex]\[ 1 + \frac{1}{r} + \frac{1}{r^2} + \ldots + \frac{1}{r^{n-1}} \][/tex]
This is also a geometric series with the first term 1 and common ratio [tex]\(\frac{1}{r}\)[/tex]. The sum of this series is:
[tex]\[ S'_n = \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} = \frac{r^n - 1}{r^{n-1}(r-1)} \][/tex]
Given [tex]\(S'_n = \frac{364}{243}\)[/tex], we have:
[tex]\[ \frac{r^n - 1}{r^{n-1}(r-1)} = \frac{364}{243} \quad \text{(2)} \][/tex]
3. Solving the equations:
From equation (1), we know that:
[tex]\[ r^n - 1 = 364 (r - 1) \][/tex]
Substituting [tex]\(r^n - 1\)[/tex] from equation (1) into equation (2):
[tex]\[ \frac{364 (r - 1)}{r^{n-1}(r - 1)} = \frac{364}{243} \][/tex]
Simplifying:
[tex]\[ \frac{364}{r^{n-1}} = \frac{364}{243} \][/tex]
[tex]\[ r^{n-1} = 243 \][/tex]
Taking the [tex]\((n-1)\)[/tex]th root:
[tex]\[ r = \sqrt[n-1]{243} \][/tex]
Now using [tex]\(r = 3\)[/tex], because [tex]\(243 = 3^5\)[/tex], we know:
[tex]\[ 3^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies n = 6 \][/tex]
4. Confirming the values:
[tex]\[ r^6 - 1 = 364(3-1) \implies 729 - 1 = 364 \times 2 \implies 728 = 728 \][/tex]
Since both conditions match perfectly, [tex]\(n = 6\)[/tex] and [tex]\(r = 3\)[/tex].
So, the [tex]\(i\)[/tex] the number of terms [tex]\(n\)[/tex], the common ratio [tex]\( r \)[/tex], and series are:
[tex]\[ \boxed{(6, 3, 1, 3, 9, 27, 81, 243)} \][/tex]
### Part 2: Solving the inequality
1. Inequality [tex]\(|3x - 5| > 7\)[/tex]:
[tex]\[ |3x - 5| > 7 \][/tex]
This implies two cases:
[tex]\[ 3x - 5 > 7 \quad \text{or} \quad 3x - 5 < -7 \][/tex]
2. Solving each case:
- Case 1: [tex]\(3x - 5 > 7\)[/tex]
[tex]\[ 3x > 12 \][/tex]
[tex]\[ x > 4 \][/tex]
- Case 2: [tex]\(3x - 5 < -7\)[/tex]
[tex]\[ 3x < -2 \][/tex]
[tex]\[ x < -\frac{2}{3} \][/tex]
So, the solution for [tex]\(|3x - 5| > 7\)[/tex] is:
[tex]\[ x > 4 \quad \text{or} \quad x < -\frac{2}{3} \][/tex]
This forms two separate ranges of [tex]\(x\)[/tex]:
[tex]\( (-\infty, -\frac{2}{3}) \cup (4, \infty) \)[/tex]
### Infinite series with [tex]\(x\)[/tex] satisfies [tex]\(|3x - 5| > 7\)[/tex]:
For [tex]\(x \in (-\infty, -\frac{2}{3}) \cup (4, \infty)\)[/tex], the value of [tex]\(x\)[/tex] would be infinite, therefore making the series infinite.
Thus, values of [tex]\( x \)[/tex] that satisfy the condition [tex]\( |3x-5|>7 \)[/tex] lead to the determination that the series is infinite.
### Part 1: Solving for the geometric series
1. Sum of the first [tex]\(n\)[/tex] terms:
We know the sum of the first [tex]\(n\)[/tex] terms of a geometric series with the first term [tex]\(a\)[/tex] and common ratio [tex]\(r\)[/tex] is:
[tex]\[ S_n = \frac{a(r^n - 1)}{r - 1} \][/tex]
Given [tex]\(S_n = 364\)[/tex] and [tex]\(a = 1\)[/tex], we have:
[tex]\[ \frac{r^n - 1}{r - 1} = 364 \quad \text{(1)} \][/tex]
2. Sum of the reciprocals of the terms:
The sum of the reciprocals of the series is given by:
[tex]\[ 1 + \frac{1}{r} + \frac{1}{r^2} + \ldots + \frac{1}{r^{n-1}} \][/tex]
This is also a geometric series with the first term 1 and common ratio [tex]\(\frac{1}{r}\)[/tex]. The sum of this series is:
[tex]\[ S'_n = \frac{1 - \left(\frac{1}{r}\right)^n}{1 - \frac{1}{r}} = \frac{r^n - 1}{r^{n-1}(r-1)} \][/tex]
Given [tex]\(S'_n = \frac{364}{243}\)[/tex], we have:
[tex]\[ \frac{r^n - 1}{r^{n-1}(r-1)} = \frac{364}{243} \quad \text{(2)} \][/tex]
3. Solving the equations:
From equation (1), we know that:
[tex]\[ r^n - 1 = 364 (r - 1) \][/tex]
Substituting [tex]\(r^n - 1\)[/tex] from equation (1) into equation (2):
[tex]\[ \frac{364 (r - 1)}{r^{n-1}(r - 1)} = \frac{364}{243} \][/tex]
Simplifying:
[tex]\[ \frac{364}{r^{n-1}} = \frac{364}{243} \][/tex]
[tex]\[ r^{n-1} = 243 \][/tex]
Taking the [tex]\((n-1)\)[/tex]th root:
[tex]\[ r = \sqrt[n-1]{243} \][/tex]
Now using [tex]\(r = 3\)[/tex], because [tex]\(243 = 3^5\)[/tex], we know:
[tex]\[ 3^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies n = 6 \][/tex]
4. Confirming the values:
[tex]\[ r^6 - 1 = 364(3-1) \implies 729 - 1 = 364 \times 2 \implies 728 = 728 \][/tex]
Since both conditions match perfectly, [tex]\(n = 6\)[/tex] and [tex]\(r = 3\)[/tex].
So, the [tex]\(i\)[/tex] the number of terms [tex]\(n\)[/tex], the common ratio [tex]\( r \)[/tex], and series are:
[tex]\[ \boxed{(6, 3, 1, 3, 9, 27, 81, 243)} \][/tex]
### Part 2: Solving the inequality
1. Inequality [tex]\(|3x - 5| > 7\)[/tex]:
[tex]\[ |3x - 5| > 7 \][/tex]
This implies two cases:
[tex]\[ 3x - 5 > 7 \quad \text{or} \quad 3x - 5 < -7 \][/tex]
2. Solving each case:
- Case 1: [tex]\(3x - 5 > 7\)[/tex]
[tex]\[ 3x > 12 \][/tex]
[tex]\[ x > 4 \][/tex]
- Case 2: [tex]\(3x - 5 < -7\)[/tex]
[tex]\[ 3x < -2 \][/tex]
[tex]\[ x < -\frac{2}{3} \][/tex]
So, the solution for [tex]\(|3x - 5| > 7\)[/tex] is:
[tex]\[ x > 4 \quad \text{or} \quad x < -\frac{2}{3} \][/tex]
This forms two separate ranges of [tex]\(x\)[/tex]:
[tex]\( (-\infty, -\frac{2}{3}) \cup (4, \infty) \)[/tex]
### Infinite series with [tex]\(x\)[/tex] satisfies [tex]\(|3x - 5| > 7\)[/tex]:
For [tex]\(x \in (-\infty, -\frac{2}{3}) \cup (4, \infty)\)[/tex], the value of [tex]\(x\)[/tex] would be infinite, therefore making the series infinite.
Thus, values of [tex]\( x \)[/tex] that satisfy the condition [tex]\( |3x-5|>7 \)[/tex] lead to the determination that the series is infinite.
