Let `{a_n}(ngeq1)` be a sequence such that `a_1=1,a n d3a_(n+1)-3a_n=1fora l lngeq1.` Then find the value of `a_(2002.)`

Explanation

To solve the problem, we need to find the value of \( a_{2002} \) given the recurrence relation and the initial condition. Let's break it down step by step.

### Step 1: Understand the recurrence relation
We are given:
- \( a_1 = 1 \)
- The recurrence relation: \( 3a_{n+1} - 3a_n = 1 \)

We can simplify this to:
\[ a_{n+1} - a_n = \frac{1}{3} \]

### Step 2: Find a general formula for \( a_n \)
From the simplified recurrence relation, we see that the difference between consecutive terms is constant:
\[ a_{n+1} = a_n + \frac{1}{3} \]

Starting from \( a_1 \):
- \( a_2 = a_1 + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3} \)
- \( a_3 = a_2 + \frac{1}{3} = \frac{4}{3} + \frac{1}{3} = \frac{5}{3} \)
- \( a_4 = a_3 + \frac{1}{3} = \frac{5}{3} + \frac{1}{3} = \frac{6}{3} = 2 \)

### Step 3: Establish a pattern
From the calculations, we can see that:
- \( a_1 = 1 \)
- \( a_2 = \frac{4}{3} \)
- \( a_3 = \frac{5}{3} \)
- \( a_4 = 2 \)

Continuing this pattern, we can express \( a_n \) in terms of \( n \):
\[ a_n = a_1 + (n-1) \cdot \frac{1}{3} = 1 + \frac{n-1}{3} \]

### Step 4: Write the formula for \( a_n \)
Thus, we have:
\[ a_n = 1 + \frac{n-1}{3} = \frac{3 + n - 1}{3} = \frac{n + 2}{3} \]

### Step 5: Calculate \( a_{2002} \)
Now we can find \( a_{2002} \):
\[ a_{2002} = \frac{2002 + 2}{3} = \frac{2004}{3} = 668 \]

### Final Answer
Therefore, the value of \( a_{2002} \) is:
\[ \boxed{668} \]