Let [x] = greatest integer `le x`. Suppose roots of the quadratic equation `x^(2)+2(b-3)x+9=0` lie between `-6 and 1`. Suppose `(1)/(2),(1)/(h_(1)),(1)/(h_(2)),…,(1)/(h_(20)),(1)/([b])` are in A.P. and `2,a_(1),a_(2),…,a_(20)`,[b] are also in A.P., then `a_(3)h_(18)=`

Answer

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Determine the range of \( b \)
We start with the quadratic equation:
\[ x^2 + 2(b-3)x + 9 = 0 \]
The roots of this equation must lie between \(-6\) and \(1\). For the roots to lie in this interval, we need to ensure that the function values at these points are positive.

1. **Evaluate \( f(1) \)**:
\[ f(1) = 1^2 + 2(b-3)(1) + 9 > 0 \]
Simplifying this gives:
\[ 1 + 2b - 6 + 9 > 0 \implies 2b + 4 > 0 \implies b > -2 \]

2. **Evaluate \( f(-6) \)**:
\[ f(-6) = (-6)^2 + 2(b-3)(-6) + 9 > 0 \]
Simplifying this gives:
\[ 36 - 12(b-3) + 9 > 0 \implies 36 - 12b + 36 > 0 \implies 72 - 12b > 0 \implies b < 6 \]

Thus, the range of \( b \) is:
\[ -2 < b < 6 \]

### Step 2: Find the greatest integer less than or equal to \( b \)
The greatest integer less than or equal to \( b \) can take values from \(-2\) to \(6\). The possible integers are:
\[ -2, -1, 0, 1, 2, 3, 4, 5, 6 \]
The greatest integer in this range is:
\[ [b] = 6 \]

### Step 3: Set up the Arithmetic Progression (AP)
We have the terms:
\[ \frac{1}{2}, \frac{1}{h_1}, \frac{1}{h_2}, \ldots, \frac{1}{h_{20}}, \frac{1}{[b]} = \frac{1}{6} \]
These terms are in AP. The first term \( a = \frac{1}{2} \) and the last term is \( \frac{1}{6} \).

Using the formula for the \( n \)-th term of an AP:
\[ T_n = a + (n-1)d \]
where \( n = 22 \) (20 terms of \( h \) plus the first and last terms). Therefore:
\[ \frac{1}{6} = \frac{1}{2} + 21d \]
Rearranging gives:
\[ 21d = \frac{1}{6} - \frac{1}{2} = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \]
Thus:
\[ d = -\frac{1}{63} \]

### Step 4: Find \( h_{18} \)
The term \( h_{18} \) is the 18th term in the sequence:
\[ h_{18} = a + (18-1)d = \frac{1}{2} + 17 \left(-\frac{1}{63}\right) \]
Calculating this gives:
\[ h_{18} = \frac{1}{2} - \frac{17}{63} = \frac{31.5 - 17}{63} = \frac{14.5}{63} = \frac{29}{126} \]

### Step 5: Set up the second AP
The terms \( 2, a_1, a_2, \ldots, a_{20}, [b] \) are also in AP. The first term \( a = 2 \) and the last term is \( 6 \).

Using the same formula:
\[ 6 = 2 + 21d' \]
Rearranging gives:
\[ 21d' = 6 - 2 = 4 \implies d' = \frac{4}{21} \]

### Step 6: Find \( a_3 \)
Using the formula for the \( n \)-th term:
\[ a_3 = 2 + (3-1)d' = 2 + 2 \left(\frac{4}{21}\right) = 2 + \frac{8}{21} = \frac{42 + 8}{21} = \frac{50}{21} \]

### Step 7: Calculate \( a_3 h_{18} \)
Now we need to find:
\[ a_3 h_{18} = \left(\frac{50}{21}\right) \left(\frac{29}{126}\right) = \frac{1450}{2646} \]
Simplifying this fraction:
\[ \frac{1450}{2646} = \frac{725}{1323} \]

### Final Answer
Thus, the final answer is:
\[ \boxed{\frac{725}{1323}} \]