Find the 20th term of an A.P. whose 5th term is 15 and the sum of its 3rd and 8th terms is 34.

Answer

To find the 20th term of an arithmetic progression (A.P.) where the 5th term is 15 and the sum of the 3rd and 8th terms is 34, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Terms of A.P.:**
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Using the Information Given:**
We know:
- The 5th term \( T_5 = 15 \)
- The sum of the 3rd and 8th terms \( T_3 + T_8 = 34 \)

3. **Setting Up the Equations:**
From the 5th term:
\[ T_5 = a + (5-1)d = a + 4d = 15 \quad \text{(Equation 1)} \]

From the sum of the 3rd and 8th terms:
\[ T_3 = a + (3-1)d = a + 2d \]
\[ T_8 = a + (8-1)d = a + 7d \]
Therefore,
\[ T_3 + T_8 = (a + 2d) + (a + 7d) = 2a + 9d = 34 \quad \text{(Equation 2)} \]

4. **Solving the Equations:**
We have two equations:
- \( a + 4d = 15 \) (Equation 1)
- \( 2a + 9d = 34 \) (Equation 2)

From Equation 1, we can express \( a \) in terms of \( d \):
\[ a = 15 - 4d \quad \text{(Substituting into Equation 2)} \]

Substitute \( a \) into Equation 2:
\[ 2(15 - 4d) + 9d = 34 \]
Simplifying:
\[ 30 - 8d + 9d = 34 \]
\[ 30 + d = 34 \]
\[ d = 34 - 30 = 4 \]

5. **Finding \( a \):**
Substitute \( d = 4 \) back into Equation 1:
\[ a + 4(4) = 15 \]
\[ a + 16 = 15 \]
\[ a = 15 - 16 = -1 \]

6. **Finding the 20th Term:**
Now that we have \( a = -1 \) and \( d = 4 \), we can find the 20th term:
\[ T_{20} = a + (20-1)d = -1 + 19 \cdot 4 \]
\[ T_{20} = -1 + 76 = 75 \]

### Final Answer:
The 20th term of the A.P. is **75**.