Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4
The sum of all numbers

Options

A. exceeds 1600
B. is less than 1500
C. lies between 1300 and 1500
D. none of these

Answer

B

Explanation

Let m and (m+1) be the removed numbers from 1,2,…,n.
Then, the sum of the remaining numbers is n(n+1)/2-(2m+1).
From given condition,
`105/4=((n(n+1))/2-(2m+1))/((n-2))`
or `2n^(2)-103n-8m+206`=0
Since n and m are integers, so n must be even. Let n=2k. Then,
`m=(4k^(2)+103(1-k))/4`
Since m is an integer, then 1-k must be divisble by 4. Let k=1+4t. Then we get n=8t+2 and `m=16t^(2)-95t+1`. Now,
`1lemltn`
`rArr1le16t^(2)-95t+1lt8t+2`
Solving, we get t=6. Hence,
n=50 and m=7
Hence, the removed numbers are 7 and 8. Also, sm of all numbers is 50(50+1)/2=1275.