Find three numbers in A.P.
(i) Whose sum is 21 and product is 315.
(ii) whose sum is 24 and product is 440.

Answer

To solve the problem of finding three numbers in arithmetic progression (A.P.) for the given conditions, we can follow these steps:

### Part (i): Sum is 21 and Product is 315

1. **Define the Numbers**:
Let the three numbers in A.P. be \( a - d \), \( a \), and \( a + d \).

2. **Set Up the Equations**:
From the problem, we know:
- The sum of the numbers:
\[ (a - d) + a + (a + d) = 21 \]
Simplifying this gives:
\[ 3a = 21 \implies a = 7 \]

- The product of the numbers:
\[ (a - d) \cdot a \cdot (a + d) = 315 \]
Substituting \( a = 7 \):
\[ (7 - d) \cdot 7 \cdot (7 + d) = 315 \]

3. **Simplify the Product Equation**:
Expanding the product:
\[ 7(7^2 - d^2) = 315 \]
This simplifies to:
\[ 7(49 - d^2) = 315 \]
Dividing both sides by 7:
\[ 49 - d^2 = 45 \implies d^2 = 4 \implies d = 2 \text{ or } d = -2 \]

4. **Find the Numbers**:
Using \( d = 2 \):
- The numbers are \( 7 - 2 = 5 \), \( 7 \), and \( 7 + 2 = 9 \).
Thus, the three numbers are \( 5, 7, 9 \).

### Part (ii): Sum is 24 and Product is 440

1. **Define the Numbers**:
Again, let the three numbers be \( a - d \), \( a \), and \( a + d \).

2. **Set Up the Equations**:
- The sum of the numbers:
\[ (a - d) + a + (a + d) = 24 \]
Simplifying gives:
\[ 3a = 24 \implies a = 8 \]

- The product of the numbers:
\[ (a - d) \cdot a \cdot (a + d) = 440 \]
Substituting \( a = 8 \):
\[ (8 - d) \cdot 8 \cdot (8 + d) = 440 \]

3. **Simplify the Product Equation**:
Expanding the product:
\[ 8(8^2 - d^2) = 440 \]
This simplifies to:
\[ 8(64 - d^2) = 440 \]
Dividing both sides by 8:
\[ 64 - d^2 = 55 \implies d^2 = 9 \implies d = 3 \text{ or } d = -3 \]

4. **Find the Numbers**:
Using \( d = 3 \):
- The numbers are \( 8 - 3 = 5 \), \( 8 \), and \( 8 + 3 = 11 \).
Thus, the three numbers are \( 5, 8, 11 \).

### Final Answers:
- For part (i): The numbers are \( 5, 7, 9 \).
- For part (ii): The numbers are \( 5, 8, 11 \).