The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Options

A. 1, 7, 13
B. 1, 4 , 7
C. 2, 8, 14
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Define the terms of the A.P.
Let the three terms of the arithmetic progression (A.P.) be:
- First term: \( a - d \)
- Second term: \( a \)
- Third term: \( a + d \)

### Step 2: Set up the equation for the sum of the terms.
According to the problem, the sum of these three terms is 21:
\[ (a - d) + a + (a + d) = 21 \]
This simplifies to:
\[ 3a = 21 \]
From this, we can solve for \( a \):
\[ a = \frac{21}{3} = 7 \]

### Step 3: Substitute \( a \) back into the terms.
Now that we have \( a \), we can express the three terms as:
- First term: \( 7 - d \)
- Second term: \( 7 \)
- Third term: \( 7 + d \)

### Step 4: Set up the equation for the product of the first and third terms.
The problem states that the product of the first and third terms exceeds the second term by 6:
\[ (7 - d)(7 + d) = 7 + 6 \]
This simplifies to:
\[ (7 - d)(7 + d) = 13 \]

### Step 5: Expand and simplify the equation.
Using the difference of squares:
\[ 49 - d^2 = 13 \]
Now, rearranging gives:
\[ 49 - 13 = d^2 \]
\[ 36 = d^2 \]

### Step 6: Solve for \( d \).
Taking the square root of both sides, we find:
\[ d = 6 \quad \text{or} \quad d = -6 \]

### Step 7: Find the three terms for both values of \( d \).
1. If \( d = 6 \):
- First term: \( 7 - 6 = 1 \)
- Second term: \( 7 \)
- Third term: \( 7 + 6 = 13 \)
- The terms are \( 1, 7, 13 \).

2. If \( d = -6 \):
- First term: \( 7 - (-6) = 13 \)
- Second term: \( 7 \)
- Third term: \( 7 + (-6) = 1 \)
- The terms are \( 13, 7, 1 \).

### Conclusion
The three terms of the A.P. are \( 1, 7, 13 \) or \( 13, 7, 1 \).

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