Two arithmetic progressions have the same numbers. The reatio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of first progression is equal to 4. The ratio of the sum of the n terms of the first progression to the sum of the n terms of teh first progression to the sum of the n terms of the second progerssion is equal to 2.
The ratio of their first term is

Answer

To solve the problem, we need to analyze the given information about the two arithmetic progressions (APs) and derive the necessary equations step by step.

### Step-by-Step Solution:

1. **Define the Terms of the APs:**
- Let the first term of the first AP be \( a \) and its common difference be \( d \).
- Let the first term of the second AP be \( b \) and its common difference be \( e \).

2. **Last Terms of the APs:**
- The last term of the first AP (for \( n \) terms) is given by:
\[ L_1 = a + (n-1)d \]
- The last term of the second AP is given by:
\[ L_2 = b + (n-1)e \]

3. **Set Up the Ratios:**
- According to the problem, we have:
\[ \frac{L_1}{b} = \frac{L_2}{a} = 4 \]
- This gives us two equations:
\[ \frac{a + (n-1)d}{b} = 4 \quad \text{(1)} \]
\[ \frac{b + (n-1)e}{a} = 4 \quad \text{(2)} \]

4. **Sum of the First n Terms:**
- The sum of the first \( n \) terms of the first AP is:
\[ S_1 = \frac{n}{2} \times (2a + (n-1)d) \]
- The sum of the first \( n \) terms of the second AP is:
\[ S_2 = \frac{n}{2} \times (2b + (n-1)e) \]
- We are given that:
\[ \frac{S_1}{S_2} = 2 \]
- This leads to:
\[ \frac{2a + (n-1)d}{2b + (n-1)e} = 2 \quad \text{(3)} \]

5. **Substituting and Rearranging:**
- From equation (1):
\[ a + (n-1)d = 4b \implies (n-1)d = 4b - a \quad \text{(4)} \]
- From equation (2):
\[ b + (n-1)e = 4a \implies (n-1)e = 4a - b \quad \text{(5)} \]

6. **Substituting into the Sum Ratio:**
- Substitute equations (4) and (5) into equation (3):
\[ \frac{2a + (4b - a)}{2b + (4a - b)} = 2 \]
- This simplifies to:
\[ \frac{a + 4b}{4a + b} = 2 \]
- Cross-multiplying gives:
\[ a + 4b = 8a + 2b \implies 6b = 7a \implies \frac{a}{b} = \frac{6}{7} \quad \text{(6)} \]

7. **Finding the Ratio of First Terms:**
- The ratio of the first terms of the two APs is:
\[ \frac{a}{b} = \frac{6}{7} \]
- Therefore, the ratio of their first terms is \( \frac{6}{7} \).

### Final Answer:
The ratio of the first terms of the two arithmetic progressions is \( \frac{6}{7} \).