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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 7,8Let p,p+1 be removed numbers from 1,2,3,...n thensum of the remaining numbers=n(n+1)2−(2p+1)∴1054=n(n+1)2−(2p+1)On evaluation we get 2n2−103n−8p+206=0Since n and p are integers so ′n′ must be even.Let n=2r We get p=4r2+(103)(1−r)4Since ′p′ is an integer so (1−r) must be divisible by 4.Let r=1+4t we get n=2+8t and p=16t2−95t+1Now 1≤p<n⇒1≤16t2−95t+1<8t+2On solving the equation16t2−95t+1<8t+2or16t2−103t−1<0 using the formula −b±√b2−4ac2awe get t=6.18 or −0.25As t is an integer we take t=6From above, n=2+8t=2+8×7=2+48=50 and p=16t2−95t+1=1662−95×6+1=7∴n=50 and p=7Hence the removed numbers are 7,8 (from above)

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