If the coefficients of the rth, `(r+1)t h ,(r+2)t h` terms is the expansion of `(1+x)^(14)` are in A.P, then the largest value of `r` is.

Explanation

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

### Step 1: Identify the coefficients
The general term \( T_k \) in the expansion of \( (1+x)^{14} \) is given by:
\[ T_k = \binom{14}{k-1} x^{k-1} \]
Thus, the coefficients of the \( r \)-th, \( (r+1) \)-th, and \( (r+2) \)-th terms are:
- Coefficient of \( r \)-th term: \( \binom{14}{r-1} \)
- Coefficient of \( (r+1) \)-th term: \( \binom{14}{r} \)
- Coefficient of \( (r+2) \)-th term: \( \binom{14}{r+1} \)

### Step 2: Set up the A.P. condition
For these coefficients to be in A.P., the following condition must hold:
\[ 2 \cdot \binom{14}{r} = \binom{14}{r-1} + \binom{14}{r+1} \]

### Step 3: Use Pascal's identity
Using Pascal's identity, we know that:
\[ \binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k} \]
Applying this to our equation:
\[ \binom{14}{r-1} + \binom{14}{r+1} = \binom{15}{r} \]
Thus, we can rewrite our A.P. condition as:
\[ 2 \cdot \binom{14}{r} = \binom{15}{r} \]

### Step 4: Express \( \binom{15}{r} \) in terms of \( \binom{14}{r} \)
Using the relationship:
\[ \binom{15}{r} = \binom{14}{r} + \binom{14}{r-1} \]
We can substitute this into our equation:
\[ 2 \cdot \binom{14}{r} = \binom{14}{r} + \binom{14}{r-1} \]
This simplifies to:
\[ \binom{14}{r} = \binom{14}{r-1} \]

### Step 5: Solve the equation
The equation \( \binom{14}{r} = \binom{14}{r-1} \) implies that:
\[ r = 7 \]
or \( r = 14 - r + 1 \) leading to \( r = 8 \).

### Step 6: Find the largest value of \( r \)
The possible values of \( r \) are \( 7 \) and \( 8 \). Therefore, the largest value of \( r \) is:
\[ \boxed{9} \]