The 18th term of an A.P. exceeds its 12th term by 24. Find the common difference.

Answer

To find the common difference of the arithmetic progression (A.P.) where the 18th term exceeds the 12th term by 24, we can follow these steps:

### Step 1: Understand the terms of the A.P.
The nth term of an A.P. can be expressed as:
\[ A_n = A + (n-1)D \]
where \( A \) is the first term, \( D \) is the common difference, and \( n \) is the term number.

### Step 2: Write the expressions for the 18th and 12th terms.
For the 18th term:
\[ A_{18} = A + (18-1)D = A + 17D \]

For the 12th term:
\[ A_{12} = A + (12-1)D = A + 11D \]

### Step 3: Set up the equation based on the given information.
According to the problem, the 18th term exceeds the 12th term by 24:
\[ A_{18} = A_{12} + 24 \]

Substituting the expressions we found:
\[ A + 17D = (A + 11D) + 24 \]

### Step 4: Simplify the equation.
Now, simplify the equation:
\[ A + 17D = A + 11D + 24 \]

Subtract \( A \) from both sides:
\[ 17D = 11D + 24 \]

### Step 5: Solve for the common difference \( D \).
Now, subtract \( 11D \) from both sides:
\[ 17D - 11D = 24 \]
\[ 6D = 24 \]

Now, divide both sides by 6:
\[ D = \frac{24}{6} = 4 \]

### Conclusion
The common difference \( D \) of the arithmetic progression is:
\[ D = 4 \]

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