Determine the A.P whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

Answer

To determine the arithmetic progression (A.P.) whose 3rd term is 16 and the 7th term exceeds the 5th term by 12, we can follow these steps:

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

The \( n \)-th term of an A.P. is given by the formula:
\[ A_n = A + (n - 1)D \]

### Step 2: Write the equations for the given terms.
1. The 3rd term \( A_3 \) is given as 16:
\[ A_3 = A + (3 - 1)D = A + 2D = 16 \quad \text{(Equation 1)} \]

2. The 7th term \( A_7 \) exceeds the 5th term \( A_5 \) by 12:
\[ A_7 - A_5 = 12 \]
Using the formula for the terms:
\[ A + (7 - 1)D - (A + (5 - 1)D = 12 \]
Simplifying this, we get:
\[ A + 6D - (A + 4D) = 12 \]
This simplifies to:
\[ 6D - 4D = 12 \implies 2D = 12 \quad \text{(Equation 2)} \]

### Step 3: Solve for \( D \).
From Equation 2:
\[ 2D = 12 \implies D = \frac{12}{2} = 6 \]

### Step 4: Substitute \( D \) back into Equation 1 to find \( A \).
Substituting \( D = 6 \) into Equation 1:
\[ A + 2(6) = 16 \]
This simplifies to:
\[ A + 12 = 16 \implies A = 16 - 12 = 4 \]

### Step 5: Write the A.P.
Now that we have \( A = 4 \) and \( D = 6 \), we can write the A.P.:
- First term: \( A = 4 \)
- Second term: \( 4 + 6 = 10 \)
- Third term: \( 10 + 6 = 16 \)
- Fourth term: \( 16 + 6 = 22 \)
- Fifth term: \( 22 + 6 = 28 \)
- Sixth term: \( 28 + 6 = 34 \)
- Seventh term: \( 34 + 6 = 40 \)

Thus, the A.P. is:
\[ 4, 10, 16, 22, 28, 34, 40, \ldots \]

### Final Answer:
The required A.P. is \( 4, 10, 16, 22, 28, 34, 40, \ldots \)

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