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Question

If x1,x2,x3.........x4001 are in an AP such that 1x1x2+1x2x3+........+1x4000x4001=10 and x2+x4000=50 then |x1−x4001|=

A
20
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B
30
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C
40
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D
100
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Solution

The correct option is B 30
x1,x2,x3,....,x4001 are in AP
then, 1x1,1x2,1x3,....,1x4001 are in HP
Given 1x1x2+1x2x3+1x3x4+.....+1x4000x4001=10
Dividing and multiplying by common difference d
1x1x2+1x2x3+1x3x4+.....+1x4000x4001=10ddx1x2+ddx2x3+ddx3x4+.....+ddx4000x4001=101d[x2x1x1x2+x3x2x2x3+x4x3x3x4+.....+x4001x4000x4000x4001]=101d[1x11x2+1x21x3+1x3....+1x40001x4001]=101d[1x11x4001]=101dx4001x1x4001x1=10(1)

an=a+(n1)d nth term formula
x4001=x1+(40011)dx4001x14000=d(2)
From (1),(2)
d×4000d=10(x4001x1)400=x4001x1(x1x4001)2=(x1+x4001)24x1x4001(x1x4001)2=(50)24×400(x1x4001)2=25001600=900(x1x4001)=±900=±30|x1x4001|=30

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