Problem: Let a1,a2,… and b1,b2,… be arithmetic progressions such that a1=25, b1=75 and a100+b100=100. Find the sum of the first one hundred terms of the progression a1+b1,a2+b2,…
Answer Choices:
A. 0
B. 100
C. 10000
D. 505000
E. not enough information given to solve the problem
Solution:
Let d denote the common difference of the progression a1+b1,a2+b2, … Then 99d=(a100+b100)−(a1+b1)=0; Thus d=0, and 100(a1+b1)=10,000 is the desired sum.