If the expansion of `(x+y)^(m)` has 3 consecutive coefficients in `A.P.` then smallest possible value of m can be (here `m in N)` is

Options

A. `9`
B. `7`
C. `2`
D. `14`

Explanation

To solve the problem, we need to find the smallest possible value of \( m \) such that the coefficients of \( (x+y)^m \) corresponding to three consecutive terms are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding Coefficients in A.P.**:
The coefficients of \( (x+y)^m \) are given by \( \binom{m}{r} \), \( \binom{m}{r+1} \), and \( \binom{m}{r+2} \). For these coefficients to be in A.P., the following condition must hold:
\[ 2 \binom{m}{r+1} = \binom{m}{r} + \binom{m}{r+2} \]

2. **Using the Binomial Coefficient Formula**:
We can express the binomial coefficients using the formula:
\[ \binom{m}{r} = \frac{m!}{r!(m-r)!} \]
\[ \binom{m}{r+1} = \frac{m!}{(r+1)!(m-r-1)!} \]
\[ \binom{m}{r+2} = \frac{m!}{(r+2)!(m-r-2)!} \]

3. **Substituting into the A.P. Condition**:
Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{m!}{(r+1)!(m-r-1)!} = \frac{m!}{r!(m-r)!} + \frac{m!}{(r+2)!(m-r-2)!} \]

4. **Cancelling \( m! \)**:
Dividing through by \( m! \) (assuming \( m \neq 0 \)):
\[ 2 \cdot \frac{1}{(r+1)(m-r-1)!} = \frac{1}{r!(m-r)!} + \frac{1}{(r+2)(m-r-2)!} \]

5. **Finding a Common Denominator**:
The common denominator for the right-hand side is \( r!(m-r)! (r+2)(m-r-2)! \). Reorganizing gives:
\[ 2(m-r)! = r!(m-r-1)! + (r+2)(m-r-2)! \]

6. **Rearranging the Equation**:
Rearranging leads to:
\[ 2(m-r) = (m-r)(m-r-1) + (r+2) \]

7. **Solving the Quadratic Equation**:
After simplification, we obtain a quadratic equation in \( r \):
\[ 4 = (m+2)(m+1)/(r+2)(m-r) \]
Cross-multiplying leads to:
\[ 4r^2 + (8 - 4m)r + (m^2 - 5m + 2) = 0 \]

8. **Finding Discriminant**:
The discriminant of this quadratic must be a perfect square for \( r \) to be an integer:
\[ D = (8 - 4m)^2 - 4 \cdot 4 \cdot (m^2 - 5m + 2) \]

9. **Setting Conditions for Perfect Square**:
Solving for \( m \) gives us conditions that lead us to test integer values for \( m \). After testing possible values, we find that \( m = 7 \) yields integer solutions for \( r \).

10. **Conclusion**:
Thus, the smallest possible value of \( m \) such that the coefficients are in A.P. is:
\[ \boxed{7} \]