InterviewSolution
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InterviewSolution
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Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4 The sum of all numbersA. less than 1000B. lies between 1200 to 1500C. greater than 1500D. None of these |
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Answer» Correct Answer - B Let p and `(p+1)` be removed numbers from `1,2,3,"……"n`, then Sum of the remaining numbers `=(n(n+1))/(2)-(2p+1)` From given condition, `(105)/(4)=((n(n1))/(2)-(2p+1))/((n-2))` `implies 2n^(2)-103n-8p+206=0` Since, n and p are integers, so n must be even. Let `n=2r`, we get `p=(4r^(2)+103(1-r))/(4)` Since, p is an integer, then `(1-r)` must be divisible by 4. Let `r=1+4t`, we get `n=2+8t` and `p=16t^(2)-95t+1` Now, `1le pltn` `implies 1le16t^(2)-95t+1lt8t+2` `implies t=6 " " implies n=50` and `p=7`. Sum of all numbers `=(50(50+1))/(2)=1275`. |
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