Find three numbers in A.P. whose sum is 21 and the product of last two numbers is 63.

Answer

To solve the problem of finding three numbers in an Arithmetic Progression (A.P.) whose sum is 21 and the product of the last two numbers is 63, we can follow these steps:

### Step 1: Define the three numbers in A.P.
Let the three numbers in 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 numbers.
According to the problem, the sum of these three numbers is given as 21:
\[ (a - d) + a + (a + d) = 21 \]
This simplifies to:
\[ 3a = 21 \]

### Step 3: Solve for \( a \).
To find \( a \), divide both sides of the equation by 3:
\[ a = \frac{21}{3} = 7 \]

### Step 4: Set up the equation for the product of the last two numbers.
The problem states that the product of the last two numbers is 63:
\[ a \cdot (a + d) = 63 \]
Substituting \( a = 7 \):
\[ 7 \cdot (7 + d) = 63 \]

### Step 5: Solve for \( d \).
First, simplify the equation:
\[ 7(7 + d) = 63 \]
Dividing both sides by 7 gives:
\[ 7 + d = 9 \]
Now, subtract 7 from both sides:
\[ d = 9 - 7 = 2 \]

### Step 6: Find the three numbers.
Now that we have \( a = 7 \) and \( d = 2 \), we can find the three numbers:
- First term: \( a - d = 7 - 2 = 5 \)
- Second term: \( a = 7 \)
- Third term: \( a + d = 7 + 2 = 9 \)

Thus, the three numbers in A.P. are:
\[ 5, 7, 9 \]

### Final Answer:
The three numbers in A.P. are **5, 7, and 9**.

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