Official GRE Quantitative Reasoning Book: Arithmetic Exercise #8

GRE Quant Problem Solving
1

The Answer Key provides this solution to Arithmetic #8:

  1. A(200) = A(100) + 101 + 102 + 103 + … + 200
  2. =A(100) + (100 + 1) + (100 + 2) + (100 + 3) + … + (100 + 100)
  3. =A(100) + A(100) + (100)(100)
  4. = 5,050 + 5,050 + 10,000
  5. = 20,100

I don’t understand how to:

a) get from the 2nd to 3rd line, specifically, where the “middle” 100, which I assume was factored out of (100 + 1) + (100 + 2) …, was relocated to, and

b) why the two 100s from the 200 were added together in step 2, but multiplied in step 3?

2

Hello,
So if you want to add to the A(100), you have 101, 102, 103 … 198, 199, 200

  • If you form pairs of first and last numbers - 101 + 200 = 301, 102 + 199 = 201, 103 + 198 = 301 …
    You will have approximately (200 - 101) + 1 numbers, sp that makes 50 pairs that make 301
    = 50 pairs * 301 = 15050

Now you add the A(100) + 15050 = 5050 + 15050 = 20100

If this is still unclear, there is a good explanation on Quant Flashcards - Group 1, sum of numbers in an interval - Quant Flashcards Group 1 - GregMat Course - GregMat