The difference of an arithmetic sequence is 3. What is the first term if the sum of the squares of the first 1001 terms is equal to the sum of the squares of the next 1000 terms?
I. solution. We use the well-known

$$
1+2+3+\cdots+(n-1)=\frac{n(n-1)}{2}
$$

and

$$
1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}=\frac{n(n-1)(2 n-1)}{6}
$$

identities. Using the usual notation,

$$
\begin{gathered}
S_{n}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=a_{1}^{2}+\left(a_{1}+d\right)^{2}+\left(a_{1}+2 d\right)^{2}+\cdots+\left(a_{1}+(n-1) d\right)^{2}= \\
=a_{1}^{2}+\left(a_{1}^{2}+2 a_{1} d+d^{2}\right)+\left(a_{1}^{2}+4 a_{1} d+4 d^{2}\right)+\cdots+\left(a_{1}^{2}+2(n-1) a_{1} d+(n-1)^{2} d^{2}\right)= \\
=n a_{1}^{2}+2(1+2+\cdots+(n-1)) a_{1} d+\left(1^{2}+2^{2}+\cdots+(n-1)^{2}\right) d^{2}= \\
=n a_{1}^{2}+n(n-1) a_{1} d+\frac{n(n-1)(2 n-1)}{6} d^{2}
\end{gathered}
$$

In the sequence of the problem, $d=3$, so

$$
S_{n}=n a_{1}^{2}+3 n(n-1) a_{1}+\frac{3}{2} n(n-1)(2 n-1)
$$

According to the second condition, the sum of the squares of the first 2001 elements is twice the sum of the squares of the first 1001 elements, i.e.,

$$
S_{2001}=2 S_{1001}
$$

Using (4):

$$
2001 a_{1}^{2}+12006000 a_{1}+24018003000=2002 a_{1}^{2}+6006000 a_{1}+6009003000
$$

Rearranging:

$$
a_{1}^{2}-6000000 a_{1}-18009000000=0
$$

Thus,

$$
a_{1}=-3000 \quad \text { or } \quad 6003000 .
$$

II. solution. If we notice that the conditions are clearly satisfied by the sequence

$-3000,-2997, \ldots, 3000$ (whose 1001st element is 0), we can see: it is more sensible to express the elements of the sequence in terms of the 1001st element. With this method, we do not even need the (2) identity.

$$
\begin{gathered}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{1001}^{2}= \\
=\left(a_{1001}-3000\right)^{2}+\left(a_{1001}-2997\right)^{2}+\cdots+\left(a_{1001}-3\right)^{2}+a_{1001}^{2}= \\
=1001 a_{1001}^{2}-6(1+2+\cdots+1000) a_{1001}+9\left(1^{2}+2^{2}+\cdots+1000^{2}\right)= \\
=1001 a_{1001}^{2}-3003000 a_{1001}+9\left(1^{2}+2^{2}+\ldots 1000^{2}\right)
\end{gathered}
$$

and

$$
\begin{gathered}
a_{1002}^{2}+\cdots+a_{2001}^{2}= \\
=\left(a_{1001}+3\right)^{2}+\left(a_{1001}+6\right)^{2}+\cdots+\left(a_{1001}+3000\right)^{2}= \\
=\left(a_{1001}^{2}+6 a_{1001}+9\right)+\left(a_{1001}^{2}+12 a_{1001}+36\right)+\cdots \\
\cdots+\left(a_{1001}^{2}+6000 a_{1001}+9000000\right)= \\
=1000 a_{1001}^{2}+6\left(1+2 \cdots+1000 a_{1001}\right)+9\left(1^{2}+2^{2}+\cdots+1000^{2}\right)= \\
=1000 a_{1001}^{2}+3003000 a_{1001}+9\left(1^{2}+2^{2}+\cdots+1000^{2}\right) .
\end{gathered}
$$

According to the conditions, these two sums are equal, i.e.,

$$
\begin{aligned}
& 1001 a_{1001}^{2}-3003000 a_{1001}+9\left(1^{2}+2^{2}+\cdots+1000^{2}\right)= \\
& =1000 a_{1001}^{2}+3003000 a_{1001}+9\left(1^{2}+2^{2}+\cdots+1000^{2}\right)
\end{aligned}
$$

Rearranging:

$$
a_{1001}^{2}-6006000 a_{1001}=0
$$

From this, $a_{1001}=0$ or $a_{1001}=6006000$, so $a_{1}=-3000$ or 6003000 .