Which term of A.P. 5, 15, 25,………….. will be 130 more than its 31st term ?

Answer

To solve the problem, we need to find which term of the arithmetic progression (A.P.) 5, 15, 25, ... will be 130 more than its 31st term.

### Step-by-Step Solution:

1. **Identify the first term (a) and common difference (d)**:
- The first term \( a = 5 \).
- The second term is \( 15 \), so the common difference \( d = 15 - 5 = 10 \).

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

3. **Find the 31st term (T_31)**:
- Using the formula:
\[ T_{31} = a + (31 - 1)d = 5 + (30)(10) = 5 + 300 = 305 \]

4. **Set up the equation for the nth term being 130 more than the 31st term**:
- We need to find \( n \) such that:
\[ T_n = T_{31} + 130 \]
- Substituting \( T_{31} \):
\[ T_n = 305 + 130 = 435 \]

5. **Write the equation for the nth term**:
- Using the nth term formula:
\[ T_n = a + (n - 1)d \]
- Substitute \( T_n = 435 \):
\[ 435 = 5 + (n - 1)(10) \]

6. **Solve for n**:
- Rearranging the equation:
\[ 435 - 5 = (n - 1)(10) \]
\[ 430 = (n - 1)(10) \]
- Divide both sides by 10:
\[ n - 1 = 43 \]
- Adding 1 to both sides gives:
\[ n = 44 \]

### Conclusion:
The term of the A.P. that is 130 more than its 31st term is the **44th term**.