Number of Quadratic equation with different condition
Solution 1:
Let us understand the condition $a+c=2b$, where $a,b,c$ are natural numbers. Here $2b$ is even, which can be obtained only by adding two numbers of same parity. Among first $2002$ natural numbers, there are $1001$ odd and $1001$ even numbers. Hence number of ordered pairs $(a,c)$ should be given by $$2\times \left( \binom{1001}{2}+\binom{1001}{2} \right) = 1000 \times 2002$$