In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.

Answer

To solve the problem, we need to find the 40th term of an arithmetic progression (A.P.) given that ten times the tenth term is equal to thirty times the thirtieth term. Let's denote the first term of the A.P. as \( A \) and the common difference as \( D \).

### Step-by-Step Solution:

1. **Identify the Terms**:
- The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = A + (n-1)D \]
- Therefore, the 10th term \( T_{10} \) is:
\[ T_{10} = A + (10-1)D = A + 9D \]
- The 30th term \( T_{30} \) is:
\[ T_{30} = A + (30-1)D = A + 29D \]

2. **Set Up the Equation**:
- According to the problem, we have:
\[ 10 \times T_{10} = 30 \times T_{30} \]
- Substituting the expressions for \( T_{10} \) and \( T_{30} \):
\[ 10(A + 9D) = 30(A + 29D) \]

3. **Expand and Simplify**:
- Expanding both sides:
\[ 10A + 90D = 30A + 870D \]
- Rearranging the equation:
\[ 10A + 90D - 30A - 870D = 0 \]
\[ -20A - 780D = 0 \]

4. **Solve for A**:
- Rearranging gives:
\[ 20A = -780D \]
- Dividing by 20:
\[ A = -39D \]

5. **Find the 40th Term**:
- The 40th term \( T_{40} \) is given by:
\[ T_{40} = A + (40-1)D = A + 39D \]
- Substituting \( A = -39D \):
\[ T_{40} = -39D + 39D = 0 \]

### Final Answer:
The 40th term of the A.P. is \( \boxed{0} \).