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Question

Two consecutive numbers from 1,2,3,...,n are removed. The Arithmetic mean of the remaining number is 1054 The removed numbers are

A
6,7
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B
7,8
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C
8,9
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D
9,10
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Solution

The correct option is B 7,8
Let p,p+1 be removed numbers from 1,2,3,...n then
sum of the remaining numbers=n(n+1)2(2p+1)
1054=n(n+1)2(2p+1)
On evaluation we get 2n2103n8p+206=0
Since n and p are integers so n must be even.
Let n=2r We get p=4r2+(103)(1r)4
Since p is an integer so (1r) must be divisible by 4.
Let r=1+4t we get n=2+8t and p=16t295t+1
Now 1p<n
116t295t+1<8t+2
On solving the equation16t295t+1<8t+2
or16t2103t1<0 using the formula b±b24ac2a
we get t=6.18 or 0.25
As t is an integer we take t=6
From above, n=2+8t=2+8×7=2+48=50 and
p=16t295t+1=166295×6+1=7
n=50 and p=7
Hence the removed numbers are 7,8 (from above)

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