enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?
A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100
For any EVENLY SPACED SET:Count = \(\frac{(biggest - smallest)}{(increment)}\) + 1
Average = median = \(\frac{(biggest + smallest)}{2}\)
Sum = (count)(average)
The INCREMENT is the difference between successive values.
Even numbers between 102 and 200, inclusive: Here, the integers are EVEN, so the increment = 2.
Count = \(\frac{(200-102)}{2}\) + 1 = 50.
Average = \(\frac{(200+102)}{2}\) = 151.
Sum = (50)(151) = 7750.
Alternate approach: 102, 104, 106......196, 198, 200.
The median is halfway between 102 and 200:
\(\frac{(102+200)}{2}\)= 151.
Of the 100 integers between 101 and 200, inclusive, half are odd, half are even.
Thus, in the set above, there are 25 even integers to the LEFT of 151 and 25 even integers to the RIGHT of 151, for a total of 50 even integers.
Add successive PAIRS of integers, working from the OUTSIDE IN:
102+200 = 302.
104+198 = 302.
106+196 = 302.
Notice that the sum of each pair = 302.
Since there will be 25 of these pairs, the following sum is yielded:
25*302 = 7550.
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