For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
What is the sum of all the even integers between 99 and 301 ?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150
Alternate approach:
100, 102, 104......198, 200, 202.....296, 298, 300.
The median is the value halfway between 100 and 300:
200.
Of the 100 integers between 100 and 199, inclusive, half are odd, half are even.
Thus, in the set above, there are 50 even integers to the LEFT of 200, implying that there are also 50 even integers to the RIGHT of 200.
Add successive PAIRS of integers, working from the OUTSIDE IN:
100+300 = 400.
102+298 = 400.
104+296 = 400.
Notice that the sum of each pair = 400.
Since there will be 50 of these pairs, the following sum is yielded:
50*400 = 20,000.
Adding in the median of 200, the final sum = 20,000 + 200 = 20,200.
The correct answer is
B.
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