If a> 0 and coefficient of `x^(2),x^(3),x^(4)` in the expansion of `(1 + x//a)^(6)` are in A.P., then a equals

Options

A. `(4+sqrt7)//3`
B. `(4+sqrt3)//3`
C. `2-sqrt3`
D. `2+sqrt3`

Answer

To solve the problem, we need to find the value of \( a \) such that the coefficients of \( x^2 \), \( x^3 \), and \( x^4 \) in the expansion of \( \left( 1 + \frac{x}{a} \right)^6 \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Write the General Term of the Expansion**:
The general term in the binomial expansion of \( (1 + \frac{x}{a})^6 \) is given by:
\[ T_{r+1} = \binom{6}{r} \left( \frac{x}{a} \right)^r \]
where \( \binom{6}{r} \) is the binomial coefficient.

2. **Find the Coefficients**:
- For \( x^2 \) (when \( r = 2 \)):
\[ \text{Coefficient of } x^2 = \binom{6}{2} \left( \frac{1}{a} \right)^2 = \frac{6 \times 5}{2 \times 1} \cdot \frac{1}{a^2} = \frac{15}{a^2} \]
- For \( x^3 \) (when \( r = 3 \)):
\[ \text{Coefficient of } x^3 = \binom{6}{3} \left( \frac{1}{a} \right)^3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \cdot \frac{1}{a^3} = \frac{20}{a^3} \]
- For \( x^4 \) (when \( r = 4 \)):
\[ \text{Coefficient of } x^4 = \binom{6}{4} \left( \frac{1}{a} \right)^4 = \frac{6 \times 5}{2 \times 1} \cdot \frac{1}{a^4} = \frac{15}{a^4} \]

3. **Set Up the A.P. Condition**:
The coefficients \( \frac{15}{a^2} \), \( \frac{20}{a^3} \), and \( \frac{15}{a^4} \) are in A.P. if:
\[ 2 \cdot \frac{20}{a^3} = \frac{15}{a^2} + \frac{15}{a^4} \]

4. **Simplify the Equation**:
Multiplying through by \( a^4 \) to eliminate the denominators:
\[ 2 \cdot 20a = 15a^2 + 15 \]
This simplifies to:
\[ 40a = 15a^2 + 15 \]
Rearranging gives:
\[ 15a^2 - 40a + 15 = 0 \]

5. **Solve the Quadratic Equation**:
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 15 \), \( b = -40 \), and \( c = 15 \).
\[ a = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 15 \cdot 15}}{2 \cdot 15} \]
\[ = \frac{40 \pm \sqrt{1600 - 900}}{30} = \frac{40 \pm \sqrt{700}}{30} \]
\[ = \frac{40 \pm 10\sqrt{7}}{30} = \frac{4 \pm \sqrt{7}}{3} \]

6. **Determine Valid Solutions**:
Since \( a > 0 \), we take:
\[ a = \frac{4 + \sqrt{7}}{3} \]

### Final Answer:
Thus, the value of \( a \) is:
\[ \boxed{\frac{4 + \sqrt{7}}{3}} \]