$10 \cdot 23$ If the sum of the first $3n$ positive integers is 150 more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is
(A) 300.
(B) 350.
(C) 400.
(D) 450.
(E) 600.
(21st American High School Mathematics Examination, 1970)
[Solution] Given $S_{3 n}-S_{n}=\frac{1}{2} \cdot 3 n(3 n+1)-\frac{1}{2} n(n+1)$ $=4 n^{2}+n=150$,

and $S_{4 n}=\frac{1}{2} \cdot 4 n(4 n+1)=2 \cdot n(4 n+1)$ $=2 \cdot\left(4 n^{2}+n\right)=300$.

Therefore, the answer is $(A)$.