Let `a_n` be the nth term of an A.P. and `a_7=22`, trhen the value of the common difference (d) that would make `a_3:a_7:a_(11)` greatest is:

Options

A. 4
B. 2
C. 0
D. 7

Answer

To solve the problem, we need to find the common difference \( d \) that maximizes the product \( a_3 \times a_7 \times a_{11} \) of the terms in an arithmetic progression (A.P.). Given that \( a_7 = 22 \), we can express the terms as follows:

1. **Identify the terms in the A.P.**:
- The \( n \)th term of an A.P. is given by:
\[ a_n = a + (n-1)d \]
- Therefore:
\[ a_3 = a + 2d \]
\[ a_7 = a + 6d \]
\[ a_{11} = a + 10d \]

2. **Set up the equation using the given information**:
- We know \( a_7 = 22 \):
\[ a + 6d = 22 \]

3. **Express \( a \) in terms of \( d \)**:
- Rearranging the equation gives:
\[ a = 22 - 6d \]

4. **Substitute \( a \) into the expressions for \( a_3 \) and \( a_{11} \)**:
- For \( a_3 \):
\[ a_3 = (22 - 6d) + 2d = 22 - 4d \]
- For \( a_{11} \):
\[ a_{11} = (22 - 6d) + 10d = 22 + 4d \]

5. **Set up the product**:
- We need to maximize the product:
\[ P = a_3 \times a_7 \times a_{11} = (22 - 4d) \times 22 \times (22 + 4d) \]

6. **Simplify the product**:
- The product can be rewritten as:
\[ P = 22 \times (22 - 4d)(22 + 4d) \]
- Using the difference of squares:
\[ P = 22 \times (22^2 - (4d)^2) = 22 \times (484 - 16d^2) \]
- Thus:
\[ P = 10648 - 352d^2 \]

7. **Maximize the product**:
- The term \( -352d^2 \) indicates that the product \( P \) is a downward-opening parabola. To maximize \( P \), we need to minimize \( d^2 \).
- The minimum value of \( d^2 \) is \( 0 \), which occurs when \( d = 0 \).

8. **Conclusion**:
- The value of \( d \) that maximizes the product \( a_3 \times a_7 \times a_{11} \) is:
\[ d = 0 \]