ABCDEF is regular hexagon with point O as centre. The value of `overset(rarr)AB+overset(rarr)AC+overset(rarr)AD+overset(rarr)AE+overset(rarr)AF = n xx overset(rarr)AO` is . Find n.

Explanation

To solve the problem, we need to find the value of \( n \) in the equation:

\[ \overset{\rarr}{AB} + \overset{\rarr}{AC} + \overset{\rarr}{AD} + \overset{\rarr}{AE} + \overset{\rarr}{AF} = n \cdot \overset{\rarr}{AO} \]

where \( ABCDEF \) is a regular hexagon with point \( O \) as the center.

### Step 1: Understanding the Vectors
First, we need to visualize the hexagon and the vectors. The vertices of the hexagon can be labeled as \( A, B, C, D, E, \) and \( F \), with \( O \) being the center. Each side of the hexagon is equal in length, and the angles between adjacent sides are \( 120^\circ \).

### Step 2: Expressing Each Vector
We can express the vectors in terms of the position vectors from the center \( O \):
- \( \overset{\rarr}{AB} \) is directed from \( A \) to \( B \).
- \( \overset{\rarr}{AC} \) is directed from \( A \) to \( C \).
- \( \overset{\rarr}{AD} \) is directed from \( A \) to \( D \).
- \( \overset{\rarr}{AE} \) is directed from \( A \) to \( E \).
- \( \overset{\rarr}{AF} \) is directed from \( A \) to \( F \).

### Step 3: Using Triangle Law
Using the triangle law of vector addition, we can express some of these vectors in terms of others:
1. \( \overset{\rarr}{AC} = \overset{\rarr}{AB} + \overset{\rarr}{BC} \)
2. \( \overset{\rarr}{AE} = \overset{\rarr}{AD} + \overset{\rarr}{DE} \)
3. \( \overset{\rarr}{AF} = \overset{\rarr}{AD} + \overset{\rarr}{DE} + \overset{\rarr}{EF} \)

### Step 4: Summing the Vectors
Now, we can sum the vectors:

\[ \overset{\rarr}{AB} + \overset{\rarr}{AC} + \overset{\rarr}{AD} + \overset{\rarr}{AE} + \overset{\rarr}{AF} \]

Substituting the expressions we derived:

\[ = \overset{\rarr}{AB} + (\overset{\rarr}{AB} + \overset{\rarr}{BC}) + \overset{\rarr}{AD} + (\overset{\rarr}{AD} + \overset{\rarr}{DE}) + (\overset{\rarr}{AD} + \overset{\rarr}{DE} + \overset{\rarr}{EF}) \]

### Step 5: Simplifying the Expression
Combining like terms, we get:

\[ = 3\overset{\rarr}{AB} + 2\overset{\rarr}{AD} + 2\overset{\rarr}{DE} + \overset{\rarr}{EF} \]

### Step 6: Canceling Opposite Vectors
In a regular hexagon:
- \( \overset{\rarr}{AB} \) and \( \overset{\rarr}{DE} \) are equal and opposite.
- \( \overset{\rarr}{BC} \) and \( \overset{\rarr}{EF} \) are also equal and opposite.

Thus, we can cancel these pairs:

\[ = 3\overset{\rarr}{AD} \]

### Step 7: Relating to \( \overset{\rarr}{AO} \)
Since \( O \) is the center of the hexagon, \( \overset{\rarr}{AD} \) can be expressed in terms of \( \overset{\rarr}{AO} \):

\[ \overset{\rarr}{AD} = 2 \cdot \overset{\rarr}{AO} \]

Substituting this back, we get:

\[ = 3(2 \cdot \overset{\rarr}{AO}) = 6 \cdot \overset{\rarr}{AO} \]

### Conclusion
Thus, we have:

\[ \overset{\rarr}{AB} + \overset{\rarr}{AC} + \overset{\rarr}{AD} + \overset{\rarr}{AE} + \overset{\rarr}{AF} = 6 \cdot \overset{\rarr}{AO} \]

This means \( n = 6 \).

### Final Answer
\[ n = 6 \]