If the coefficients of r,(r +1 ),(r+2) terms in `(1+x)^14` are in A.P. then r =

Options

A. 3,2
B. 5,9
C. 2,4
D. 5,3

Answer

To solve the problem, we need to find the value of \( r \) such that the coefficients of the \( r \), \( r + 1 \), and \( r + 2 \) terms in the expansion of \( (1 + x)^{14} \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The coefficient of the \( r \)-th term in the expansion of \( (1 + x)^{14} \) is given by \( \binom{14}{r} \).
- Coefficient of \( r \)-th term: \( \binom{14}{r} \)
- Coefficient of \( (r + 1) \)-th term: \( \binom{14}{r + 1} \)
- Coefficient of \( (r + 2) \)-th term: \( \binom{14}{r + 2} \)

2. **Set Up the A.P Condition**:
The coefficients are in A.P. if:
\[ 2 \cdot \binom{14}{r + 1} = \binom{14}{r} + \binom{14}{r + 2} \]

3. **Use the Property of Binomial Coefficients**:
We can express \( \binom{14}{r + 2} \) in terms of \( \binom{14}{r} \) and \( \binom{14}{r + 1} \):
\[ \binom{14}{r + 2} = \frac{14 - r}{r + 2} \cdot \binom{14}{r + 1} \]
\[ \binom{14}{r} = \frac{r + 1}{14 - r} \cdot \binom{14}{r + 1} \]

4. **Substituting into the A.P Condition**:
Substitute the expressions into the A.P condition:
\[ 2 \cdot \binom{14}{r + 1} = \binom{14}{r} + \binom{14}{r + 2} \]
becomes:
\[ 2 \cdot \binom{14}{r + 1} = \frac{r + 1}{14 - r} \cdot \binom{14}{r + 1} + \frac{14 - r}{r + 2} \cdot \binom{14}{r + 1} \]

5. **Factor Out \( \binom{14}{r + 1} \)**:
Since \( \binom{14}{r + 1} \) is common, we can factor it out:
\[ 2 = \frac{r + 1}{14 - r} + \frac{14 - r}{r + 2} \]

6. **Clear the Denominators**:
Multiply through by \( (14 - r)(r + 2) \):
\[ 2(14 - r)(r + 2) = (r + 1)(r + 2) + (14 - r)(14 - r) \]

7. **Expand and Simplify**:
Expanding both sides:
\[ 2(14r + 28 - r^2 - 2r) = r^2 + 3r + 2 + 196 - 28r + r^2 \]
Simplifying gives:
\[ 28r + 56 - 2r^2 = 2r^2 - 25r + 198 \]

8. **Rearranging the Equation**:
Rearranging terms leads to:
\[ 4r^2 - 53r + 142 = 0 \]

9. **Using the Quadratic Formula**:
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{53 \pm \sqrt{(-53)^2 - 4 \cdot 4 \cdot 142}}{2 \cdot 4} \]
\[ r = \frac{53 \pm \sqrt{2809 - 2272}}{8} \]
\[ r = \frac{53 \pm \sqrt{537}}{8} \]

10. **Finding the Values of \( r \)**:
Calculate the values:
\[ r_1 = \frac{53 + \sqrt{537}}{8}, \quad r_2 = \frac{53 - \sqrt{537}}{8} \]

### Final Answer:
The values of \( r \) are approximately \( 9 \) and \( 5 \).