$$
\begin{array}{l}
2=2 \times 1 \\
6=2(1+2) \\
12=2(1+2+3) \\
20=2(1+2+3+4) \\
30=2(1+2+3+4+5)
\end{array}
$$

The last number of the $50^{\text {th }}$ group
$$
\begin{array}{l}
=2(1+2+\ldots+50) \\
=2 \cdot \frac{1}{2} \cdot 50 \cdot(1+50)=2550
\end{array}
$$

There are 50 numbers in the $50^{\text {th }}$ group.
The first number of the $50^{\text {th }}$ group $=2550-2(50-1)=2452$
$$
\begin{array}{l}
2452+2454+\ldots+2550=50 P \\
\frac{1}{2} \cdot 50 \cdot(2452+2550)=50 P \\
P=2501
\end{array}
$$
SG. 1 Find the last number of the $50^{\text {th }}$ group.
SG. 2 Find the first number of the $50^{\text {th }}$ group.
SG. 3 Find $P$ if the sum of the numbers in the $50^{\text {th }}$ group is $50 P$.