Answer :
To solve this problem, let's consider the characteristics of an arithmetic progression and use given conditions to derive the solution.
### Step-by-Step Solution
1. Identify Initial Conditions:
- The first term of the arithmetic progression (AP) is denoted as \( a_1 = \gamma \), where \(\gamma < 0\).
- It is given that \( a_k = 0 \). This indicates that the \(k\)-th term of the AP is zero.
2. General Term of an Arithmetic Progression:
- For an AP, the general term \( a_n \) can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Here, \( d \) is the common difference between consecutive terms.
3. Use Condition \( a_k = 0 \) to Find \( d \):
- According to the given condition \( a_k = 0 \):
[tex]\[ a_k = a_1 + (k-1)d = 0 \][/tex]
- Substitute \( a_1 = \gamma \):
[tex]\[ \gamma + (k-1)d = 0 \][/tex]
- Solve for \( d \):
[tex]\[ d = -\frac{\gamma}{k-1} \][/tex]
4. Form the Series \( S_2 \) to \( S_{2k-1} \):
- We need to find the terms of the series from \( S_2 \) to \( S_{2k-1} \). Essentially, we are interested in terms from \( a_2 \) to \( a_{2k-1} \).
5. Calculate Each Term:
- For \( n = 2 \) to \( 2k-1 \), the \( n \)-th term is:
[tex]\[ a_n = a_1 + (n-1)d = \gamma + (n-1) \left( -\frac{\gamma}{k-1} \right) \][/tex]
- Simplify inside the parentheses:
[tex]\[ a_n = \gamma - \frac{\gamma (n-1)}{k-1} \][/tex]
- Factor out \(\gamma\):
[tex]\[ a_n = \gamma \left( 1 - \frac{n-1}{k-1} \right) \][/tex]
- Simplify the fraction:
[tex]\[ a_n = \gamma \left( 1 - \frac{n-1}{k-1} \right) = \gamma \cdot \frac{k-1 - (n-1)}{k-1} = \gamma \cdot \frac{k-n}{k-1} \][/tex]
6. Write the Series:
- The series \( S_2 \) to \( S_{2k-1} \) consists of the terms \( a_n \) where \( n \) ranges from 2 to \( 2k-1 \):
[tex]\[ S_2 \sim S_{2k-1} = \left\{ \gamma \cdot \frac{k-2}{k-1}, \gamma \cdot \frac{k-3}{k-1}, \gamma \cdot \frac{k-4}{k-1}, \ldots, \gamma \cdot \frac{k-(2k-1)}{k-1} \right\} \][/tex]
7. Simplify Each Term:
- Specifically breaking down a few terms:
[tex]\[ a_2 = \gamma \cdot \frac{k-2}{k-1} \][/tex]
[tex]\[ a_3 = \gamma \cdot \frac{k-3}{k-1} \][/tex]
[tex]\[ ... \][/tex]
[tex]\[ a_{2k-1} = \gamma \cdot \frac{k-(2k-1)}{k-1} = \gamma \cdot \frac{k-2k+1}{k-1} = \gamma \cdot \frac{1-k}{k-1} \][/tex]
Thus, the series \( S_2 \) to \( S_{2k-1} \) is:
[tex]\[ \left\{ \gamma \cdot \frac{k-2}{k-1}, \gamma \cdot \frac{k-3}{k-1}, \gamma \cdot \frac{k-4}{k-1}, \ldots, \gamma \cdot \frac{1-2k}{k-1} \right\} \][/tex]
### Step-by-Step Solution
1. Identify Initial Conditions:
- The first term of the arithmetic progression (AP) is denoted as \( a_1 = \gamma \), where \(\gamma < 0\).
- It is given that \( a_k = 0 \). This indicates that the \(k\)-th term of the AP is zero.
2. General Term of an Arithmetic Progression:
- For an AP, the general term \( a_n \) can be expressed as:
[tex]\[ a_n = a_1 + (n-1)d \][/tex]
Here, \( d \) is the common difference between consecutive terms.
3. Use Condition \( a_k = 0 \) to Find \( d \):
- According to the given condition \( a_k = 0 \):
[tex]\[ a_k = a_1 + (k-1)d = 0 \][/tex]
- Substitute \( a_1 = \gamma \):
[tex]\[ \gamma + (k-1)d = 0 \][/tex]
- Solve for \( d \):
[tex]\[ d = -\frac{\gamma}{k-1} \][/tex]
4. Form the Series \( S_2 \) to \( S_{2k-1} \):
- We need to find the terms of the series from \( S_2 \) to \( S_{2k-1} \). Essentially, we are interested in terms from \( a_2 \) to \( a_{2k-1} \).
5. Calculate Each Term:
- For \( n = 2 \) to \( 2k-1 \), the \( n \)-th term is:
[tex]\[ a_n = a_1 + (n-1)d = \gamma + (n-1) \left( -\frac{\gamma}{k-1} \right) \][/tex]
- Simplify inside the parentheses:
[tex]\[ a_n = \gamma - \frac{\gamma (n-1)}{k-1} \][/tex]
- Factor out \(\gamma\):
[tex]\[ a_n = \gamma \left( 1 - \frac{n-1}{k-1} \right) \][/tex]
- Simplify the fraction:
[tex]\[ a_n = \gamma \left( 1 - \frac{n-1}{k-1} \right) = \gamma \cdot \frac{k-1 - (n-1)}{k-1} = \gamma \cdot \frac{k-n}{k-1} \][/tex]
6. Write the Series:
- The series \( S_2 \) to \( S_{2k-1} \) consists of the terms \( a_n \) where \( n \) ranges from 2 to \( 2k-1 \):
[tex]\[ S_2 \sim S_{2k-1} = \left\{ \gamma \cdot \frac{k-2}{k-1}, \gamma \cdot \frac{k-3}{k-1}, \gamma \cdot \frac{k-4}{k-1}, \ldots, \gamma \cdot \frac{k-(2k-1)}{k-1} \right\} \][/tex]
7. Simplify Each Term:
- Specifically breaking down a few terms:
[tex]\[ a_2 = \gamma \cdot \frac{k-2}{k-1} \][/tex]
[tex]\[ a_3 = \gamma \cdot \frac{k-3}{k-1} \][/tex]
[tex]\[ ... \][/tex]
[tex]\[ a_{2k-1} = \gamma \cdot \frac{k-(2k-1)}{k-1} = \gamma \cdot \frac{k-2k+1}{k-1} = \gamma \cdot \frac{1-k}{k-1} \][/tex]
Thus, the series \( S_2 \) to \( S_{2k-1} \) is:
[tex]\[ \left\{ \gamma \cdot \frac{k-2}{k-1}, \gamma \cdot \frac{k-3}{k-1}, \gamma \cdot \frac{k-4}{k-1}, \ldots, \gamma \cdot \frac{1-2k}{k-1} \right\} \][/tex]
