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.