Problem: 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
Answer Choices:
A. 300
B. 350
C. 400
D. 450
E. 600
Solution:
Let Sm denote the sum of the first m positive integers. Using the formula for the sum of an arithmetic progression, S3n−Sn=3n(3n+1)/2−n(n+1)/2=4n2+n= 150∴4n2+n−150=0∴(n−6)(4n+25)=0∴n=6∴S4n=S24=12(24+1)=300.