(1/26)+(1/27)+……+(1/49)+(1/50) is included in which of the following r

Question

(1/26)+(1/27)+……+(1/49)+(1/50) is included in which of the following ranges?
 
A. (1/6)~(1/5) 
B. (1/5)~91/4) 
C. (1/4)~(1/3) 
D. (1/3)~(1/2) 
E. (1/2)~1

souvonik2k · Accepted Answer

(1/26)+(1/27)+……+(1/49)+(1/50)
There are 25 terms in this series.
Smallest sum will be in the case when all terms are equal to  \frac{1}{50} 
Smallest sum =  \frac{25}{50}  = \frac{1}{2} 
Largest sum will be in the case when all terms are equal to  \frac{1}{26} 
Largest sum = \frac{25}{26}  ~ 1
Only option E is in this range
Answer E.

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sgw131 · Answer

First you need to figure out how many numbers there are in the set. I took the 50 from 1/50 and the 26 from 1/26:

50-26+1 will give you 25 total. That is the formula for to count a given set that is inclusive.

Next, you can take the largest number in the set and assume all the numbers were that (1/26) , if you multiply 25 by 1/26 you end up with 25/26. This is greater than what our actual sum will be so lets say X = the sum of the set provided. Then X < 25/26.

You need to find the lower end of the range by assuming that all the numbers are the lowest possible value (1/50). Multiply that by 25 to get 1/2. This is the bottom end of the range so X > 1/2.

Put both of these together to get 1/2 < X < 25/26, the only answer that fits would be E.

Hopefully that makes sense.

MathRevolution · Answer

=>
1/26 > 1/50, 1/27 > 1/50, …, 1/49 > 1/50
(1/26)+(1/27)+……+(1/49)+(1/50) > 1/50 + 1/50 + … + 1/50 = (1/50)*25 = 1/2.
1/26 < 1/25, 1/27 < 1/25, … , 1/49 < 1/25, 1/50 < 1/25.
(1/26)+(1/27)+……+(1/49)+(1/50) < 1/25 + 1/25 + … + 1/25 = (1/25)*25 = 1.

Therefore, the answer is E.
Answer : E

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(1/26)+(1/27)+……+(1/49)+(1/50) is included in which of the following r

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(1/26)+(1/27)+……+(1/49)+(1/50) is included in which of the following r

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