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![[1/2+1/100] + [1/2+2/100] + [1/2+3/100] + ...... + [1/2+103/](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "[1/2+1/100] + [1/2+2/100] + [1/2+3/100] + ...... + [1/2+103/"){kind=icon}
11 Oct 2012, 20:50
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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0% (00:00) correct
100%
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wrong
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Guys plz help me with the question below
[1/2+1/100] + [1/2+2/100] + [1/2+3/100] + ...... + [1/2+103/100]
a)52
b)53
c)54
d)55
e)none of these
Thanks
[1/2+1/100] + [1/2+2/100] + [1/2+3/100] + ...... + [1/2+103/100]
a)52
b)53
c)54
d)55
e)none of these
Thanks
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11 Oct 2012, 21:41
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154238
There is something wrong in the answer choices or the question (or the answer is (E)).
Ignore the 1/2s right now.
You have 1/100 and you have 99/100 in the series. When you add them, you will get 1.
Similarly, you have 2/100 and 98/100. This will again give you 1.
This will continue till 49/100 and 51/100.
So you have a total of 49.
You are left with 5 terms: 50/100 + 100/100 + 101/100 + 102/100 + 103/100 which add to apprx 4.5
Now you have a 103 times 1/2 which adds up to 51.5
So the total is 49 + 4.5 + 51.5 = 105 apprx
11 Oct 2012, 21:46
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You can also use the concept of sum of first n integers here.
Sum of first n integers = n(n+1)/2
Take all 1/2s together and take out 1/100 common out of the rest of the terms.
103*(1/2) + (1/100)(1 + 2 + 3 + 4 ....+ 103)
= 103/2 + (1/100) (103*104)/2 = 105 apprx
Sum of first n integers = n(n+1)/2
Take all 1/2s together and take out 1/100 common out of the rest of the terms.
103*(1/2) + (1/100)(1 + 2 + 3 + 4 ....+ 103)
= 103/2 + (1/100) (103*104)/2 = 105 apprx
