The sum of 5th and 9th terms ofan AP is 72 and the sum of 7th and 12th terms is 97. Find the AP.

Answer

To solve the problem step by step, we will find the first term (A) and the common difference (d) of the arithmetic progression (AP) using the given conditions.

### Step 1: Define the terms of the AP
Let the first term of the AP be \( A \) and the common difference be \( d \).

### Step 2: Write the expressions for the 5th, 7th, 9th, and 12th terms
- The 5th term \( T_5 = A + (5-1)d = A + 4d \)
- The 9th term \( T_9 = A + (9-1)d = A + 8d \)
- The 7th term \( T_7 = A + (7-1)d = A + 6d \)
- The 12th term \( T_{12} = A + (12-1)d = A + 11d \)

### Step 3: Set up the equations based on the given information
From the problem, we know:
1. The sum of the 5th and 9th terms is 72:
\[ T_5 + T_9 = (A + 4d) + (A + 8d) = 2A + 12d = 72 \quad \text{(Equation 1)} \]

2. The sum of the 7th and 12th terms is 97:
\[ T_7 + T_{12} = (A + 6d) + (A + 11d) = 2A + 17d = 97 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 2A + 12d = 72 \)
2. \( 2A + 17d = 97 \)

To eliminate \( A \), we can subtract Equation 1 from Equation 2:
\[ (2A + 17d) - (2A + 12d) = 97 - 72 \]
This simplifies to:
\[ 5d = 25 \]
From this, we can find \( d \):
\[ d = \frac{25}{5} = 5 \]

### Step 5: Substitute \( d \) back to find \( A \)
Now, substitute \( d = 5 \) back into Equation 1:
\[ 2A + 12(5) = 72 \]
This simplifies to:
\[ 2A + 60 = 72 \]
Subtract 60 from both sides:
\[ 2A = 72 - 60 \]
\[ 2A = 12 \]
Now, divide by 2:
\[ A = \frac{12}{2} = 6 \]

### Step 6: Write the AP
Now that we have \( A = 6 \) and \( d = 5 \), we can write the AP:
- The first term \( A = 6 \)
- The second term \( A + d = 6 + 5 = 11 \)
- The third term \( A + 2d = 6 + 2(5) = 16 \)
- The fourth term \( A + 3d = 6 + 3(5) = 21 \)
- The fifth term \( A + 4d = 6 + 4(5) = 26 \)
- Continuing this, we have the AP: \( 6, 11, 16, 21, 26, \ldots \)

### Final Answer
The arithmetic progression (AP) is:
\[ 6, 11, 16, 21, 26, \ldots \]