Suppose n be an integer greater than 1 , let `a_n=(1)/(log_n 2002 )`
Suppose `b=a_2+a_3+a_4+a_5 and c=a_(10)+a_11+a_(12)+a_(13)+a_(14)`. Then |b-c| equals __________.

Answer

To solve the problem, we need to compute the values of \( b \) and \( c \) as defined in the question and then find \( |b - c| \).

### Step 1: Calculate \( b \)

Given:
\[ b = a_2 + a_3 + a_4 + a_5 \]
where \( a_n = \frac{1}{\log_n 2002} \).

Using the change of base formula for logarithms, we have:
\[ a_n = \frac{1}{\log_n 2002} = \log_{2002} n \]

Thus, we can rewrite \( b \) as:
\[ b = \log_{2002} 2 + \log_{2002} 3 + \log_{2002} 4 + \log_{2002} 5 \]

Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine these:
\[ b = \log_{2002} (2 \times 3 \times 4 \times 5) \]

Calculating the product:
\[ 2 \times 3 = 6, \quad 6 \times 4 = 24, \quad 24 \times 5 = 120 \]
Thus,
\[ b = \log_{2002} 120 \]

### Step 2: Calculate \( c \)

Similarly, for \( c \):
\[ c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14} \]
Using the same transformation:
\[ c = \log_{2002} 10 + \log_{2002} 11 + \log_{2002} 12 + \log_{2002} 13 + \log_{2002} 14 \]

Combining these using the logarithmic property:
\[ c = \log_{2002} (10 \times 11 \times 12 \times 13 \times 14) \]

Calculating the product:
\[ 10 \times 11 = 110, \quad 110 \times 12 = 1320, \quad 1320 \times 13 = 17160, \quad 17160 \times 14 = 240240 \]
Thus,
\[ c = \log_{2002} 240240 \]

### Step 3: Calculate \( |b - c| \)

Now we need to find:
\[ |b - c| = |\log_{2002} 120 - \log_{2002} 240240| \]

Using the property of logarithms:
\[ \log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \]
So we have:
\[ |b - c| = \left| \log_{2002} \left(\frac{120}{240240}\right) \right| \]

Calculating the fraction:
\[ \frac{120}{240240} = \frac{1}{2002} \]
Thus:
\[ |b - c| = \left| \log_{2002} \left(\frac{1}{2002}\right) \right| \]

Using the property \( \log_a \left(\frac{1}{b}\right) = -\log_a b \):
\[ |b - c| = \left| -\log_{2002} 2002 \right| = 1 \]

### Final Answer

Thus, the final answer is:
\[ \boxed{1} \]