The 5th term of an A.P. is three times the first term. Prove that its 7th term will be two times the 3rd term.

Explanation

To solve the problem, we need to prove that the 7th term of an arithmetic progression (A.P.) is twice the 3rd term, given that the 5th term is three times the first term.

### Step-by-Step Solution:

1. **Define the terms of the A.P.**:
Let the first term of the A.P. be \( A \) and the common difference be \( D \).

2. **Write the formula for the nth term**:
The nth term of an A.P. is given by the formula:
\[ A_n = A + (n - 1)D \]

3. **Find the 5th term**:
The 5th term \( A_5 \) can be calculated as:
\[ A_5 = A + (5 - 1)D = A + 4D \]

4. **Set up the equation based on the given condition**:
According to the problem, the 5th term is three times the first term:
\[ A + 4D = 3A \]

5. **Rearrange the equation**:
Rearranging the equation gives:
\[ 4D = 3A - A \]
\[ 4D = 2A \]
Dividing both sides by 2:
\[ 2D = A \quad \text{(Equation 1)} \]

6. **Find the 7th term**:
The 7th term \( A_7 \) is given by:
\[ A_7 = A + (7 - 1)D = A + 6D \]

7. **Substitute \( A \) from Equation 1 into the 7th term**:
From Equation 1, we know \( A = 2D \). Substituting this into the expression for the 7th term:
\[ A_7 = 2D + 6D = 8D \]

8. **Find the 3rd term**:
The 3rd term \( A_3 \) is given by:
\[ A_3 = A + (3 - 1)D = A + 2D \]
Substituting \( A = 2D \):
\[ A_3 = 2D + 2D = 4D \]

9. **Prove that the 7th term is twice the 3rd term**:
Now we need to check if \( A_7 = 2 \times A_3 \):
\[ A_7 = 8D \quad \text{and} \quad 2 \times A_3 = 2 \times 4D = 8D \]
Thus, we have:
\[ A_7 = 2 \times A_3 \]

### Conclusion:
We have proved that the 7th term of the A.P. is indeed twice the 3rd term.