VJesus12 wrote:The sum S of the arithmetic sequence $$a,\ a+d,\ a+2d,\ \dots\ ,\ a+(n-1)\cdot d$$ is give by $$Sn=\frac{n}{2}(2a+(n-1)\cdot d\ ).$$ What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted?
A. 4345
B. 4302
C. 4258
D. 4214
E. 4170
For any EVENLY SPACED SET:
Count = (biggest - smallest)/(increment) + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).
The INCREMENT is the difference between successive values.
Integers between 1 and 100, inclusive:
Here, the integers are CONSECUTIVE, so the increment = 1.
Count = (100-1)/1 + 1 = 100.
Average = (100 + 1)/2 = 101/2.
Sum = (100)(101/2) = 5050.
Even integers between 26 and 62, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = (62-26)/2 + 1 = 19.
Average = (62+26)/2= 44.
Sum = (19)(44) = (20-1)(44) = 880-44 = 836.
Subtracting the second sum from the first, we get:
5050 - 836 = 4214.
The correct answer is
D.
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