The Answer Key provides this solution to Arithmetic #8:

- A(200) = A(100) + 101 + 102 + 103 + … + 200
- =A(100) + (100 + 1) + (100 + 2) + (100 + 3) + … + (100 + 100)
- =A(100) + A(100) + (100)(100)
- = 5,050 + 5,050 + 10,000
- = 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?

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