G1.2 Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic sequence with common difference 1 and $a_{1}+a_{2}+a_{3}+\ldots+a_{100}=2012$. If $P=a_{2}+a_{4}+a_{6}+\ldots+a_{100}$, find the value of $P$.
Let $a_{1}=a$ then $\frac{100(2 a+99)}{2}=2012$
$$
\begin{array}{l}
2 a+99=\frac{1006}{25} \\
2 a=\frac{1006-99 \times 25}{25}=-\frac{1469}{25} \\
P=a_{2}+a_{4}+a_{6}+\ldots+a_{100}=\frac{50(a+1+a+99)}{2}=25 \times(2 a+100)=25 \times\left(100-\frac{1469}{25}\right) \\
P=2500-1469=1031 \\
\text { Method } 2 P=a_{2}+a_{4}+a_{6}+\ldots+a_{100} \\
Q=a_{1}+a_{3}+a_{3}+\ldots+a_{99} \\
P-Q=1+1+1+\ldots+1 \text { (50 terms) }=50 \\
\end{array}
$$

But since $P+Q=a_{1}+a_{2}+a_{3}+\ldots+a_{100}=2012$
$$
\therefore P=\frac{2012+50}{2}=1031
$$

Remark: the original question ...等差級數..., ... arithmetic progression...
The phrases are changed to ...等差数列.... and ...arithmetic sequence... according to the mathematics syllabus since 1999 .