Which term of the A.P. 4,11,18,25,…… is 42 more than its 25th term?

Answer

To solve the problem, we need to determine which term of the arithmetic progression (A.P.) 4, 11, 18, 25, ... is 42 more than its 25th term.

### Step-by-Step Solution:

1. **Identify the first term (a) and the common difference (d)**:
- The first term \( a = 4 \)
- The second term \( a_2 = 11 \)
- The common difference \( d = a_2 - a_1 = 11 - 4 = 7 \)

2. **Find the 25th term (\( a_{25} \)) of the A.P.**:
- The formula for the \( n \)-th term of an A.P. is given by:
\[ a_n = a + (n - 1) \cdot d \]
- For the 25th term:
\[ a_{25} = 4 + (25 - 1) \cdot 7 = 4 + 24 \cdot 7 = 4 + 168 = 172 \]

3. **Set up the equation for the term that is 42 more than the 25th term**:
- Let the term we are looking for be \( a_n \).
- According to the problem, \( a_n = a_{25} + 42 \):
\[ a_n = 172 + 42 = 214 \]

4. **Use the formula for the \( n \)-th term to find \( n \)**:
- We have:
\[ a_n = a + (n - 1) \cdot d \]
- Substituting the known values:
\[ 214 = 4 + (n - 1) \cdot 7 \]
- Rearranging gives:
\[ 214 - 4 = (n - 1) \cdot 7 \]
\[ 210 = (n - 1) \cdot 7 \]

5. **Solve for \( n - 1 \)**:
- Divide both sides by 7:
\[ n - 1 = \frac{210}{7} = 30 \]

6. **Find \( n \)**:
- Add 1 to both sides:
\[ n = 30 + 1 = 31 \]

### Conclusion:
The term of the A.P. that is 42 more than its 25th term is the **31st term**.

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