If coefficient of 9th, 10th and 11th terms of `(1+x)^(n)` are in A.P., then the value of n can be

Options

A. 15
B. 32
C. 23
D. 17

Answer

To solve the problem, we need to find the value of \( n \) such that the coefficients of the 9th, 10th, and 11th terms of the expansion of \( (1+x)^n \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (1+x)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^r \]
Therefore, the coefficients of the 9th, 10th, and 11th terms are:
- 9th term: \( T_9 = \binom{n}{8} \)
- 10th term: \( T_{10} = \binom{n}{9} \)
- 11th term: \( T_{11} = \binom{n}{10} \)

2. **Set Up the A.P. Condition**:
The coefficients are in A.P. if:
\[ 2 \cdot \binom{n}{9} = \binom{n}{8} + \binom{n}{10} \]

3. **Use the Property of Binomial Coefficients**:
We can use the property of binomial coefficients:
\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} \]
Applying this property:
- From \( \binom{n}{9} \) to \( \binom{n}{8} \):
\[ \frac{\binom{n}{8}}{\binom{n}{9}} = \frac{9}{n-8} \]
- From \( \binom{n}{10} \) to \( \binom{n}{9} \):
\[ \frac{\binom{n}{9}}{\binom{n}{10}} = \frac{10}{n-9} \]

4. **Substituting into the A.P. Condition**:
Substitute these into the A.P. condition:
\[ 2 \cdot \binom{n}{9} = \binom{n}{8} + \binom{n}{10} \]
becomes:
\[ 2 \cdot \binom{n}{9} = \frac{9}{n-8} \cdot \binom{n}{9} + \frac{10}{n-9} \cdot \binom{n}{9} \]
Dividing through by \( \binom{n}{9} \) (assuming \( \binom{n}{9} \neq 0 \)):
\[ 2 = \frac{9}{n-8} + \frac{10}{n-9} \]

5. **Finding a Common Denominator**:
The common denominator is \( (n-8)(n-9) \):
\[ 2(n-8)(n-9) = 9(n-9) + 10(n-8) \]
Expanding both sides:
\[ 2(n^2 - 17n + 72) = 9n - 81 + 10n - 80 \]
Simplifying:
\[ 2n^2 - 34n + 144 = 19n - 161 \]
Rearranging gives:
\[ 2n^2 - 53n + 305 = 0 \]

6. **Using the Quadratic Formula**:
Now we can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -53, c = 305 \):
\[ n = \frac{53 \pm \sqrt{(-53)^2 - 4 \cdot 2 \cdot 305}}{2 \cdot 2} \]
\[ n = \frac{53 \pm \sqrt{2809 - 2440}}{4} \]
\[ n = \frac{53 \pm \sqrt{369}}{4} \]

7. **Calculating the Values**:
The approximate value of \( \sqrt{369} \) is about \( 19.2 \):
\[ n = \frac{53 \pm 19.2}{4} \]
This gives us two possible values for \( n \):
\[ n_1 = \frac{72.2}{4} \approx 18.05 \quad \text{and} \quad n_2 = \frac{33.8}{4} \approx 8.45 \]

8. **Conclusion**:
Since \( n \) must be a non-negative integer, we check the possible integer values around 18.05 and 8.45. The only integer solution that satisfies the condition is \( n = 18 \).

### Final Answer:
The value of \( n \) can be \( 18 \).