In a harmonic sequence, the reciprocals of the terms form an arithmetic sequence. Let the terms of the harmonic sequence be represented as \( a_1, a_2, a_3, \ldots, a_{11} \). Then the reciprocals can be represented as:
\[ \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_{11}} \]
Let’s denote the first term of the reciprocal sequence (which is an arithmetic sequence) as \( \frac{1}{a} \) and the common difference as \( d \). Thus, the terms of the reciprocal sequence can be expressed as follows:
\[ \frac{1}{a_1} = \frac{1}{a} \ \frac{1}{a_2} = \frac{1}{a} + d \ \frac{1}{a_3} = \frac{1}{a} + 2d \ \vdots \ \frac{1}{a_{11}} = \frac{1}{a} + 10d \]
The sum of the reciprocals of 11 terms is given to be 110:
\[ \sum_{n=0}^{10} \left( \frac{1}{a} + nd \right) = 110 \]
This expands to:
\[ 11 \left( \frac{1}{a} \right) + d(0 + 1 + 2 + \ldots + 10) = 110 \]
The sum of the first 10 integers is:
\[ 0 + 1 + 2 + \ldots + 10 = \frac{10 \cdot 11}{2} = 55 \]
Putting it back into the equation gives us:
\[ 11 \left( \frac{1}{a} \right) + 55d = 110 \]
Subtracting 110 from both sides:
\[ 11 \cdot \frac{1}{a} + 55d = 110 \]
Now, we can simplify by dividing every term by 11:
\[ \frac{1}{a} + 5d = 10 \]
From here, we can express \( \frac{1}{a} \) in terms of \( d \):
\[ \frac{1}{a} = 10 - 5d \]
Next, to find the sixth term of the harmonic sequence, we need \( a_6 \):
\[ \frac{1}{a_6} = \frac{1}{a} + 5d = (10 - 5d) + 5d = 10 \]
Thus, we have:
\[ a_6 = \frac{1}{10} \]
Therefore, the sixth term of the harmonic sequence is
\[ \boxed{\frac{1}{10}} \]
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