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Question

x<1 and x,|x+1|,|x1| are in A.P. The sum of the first twenty terms of the A.P. is

A
150
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B
280
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C
350
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D
180
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Solution

The correct option is B 350
x<1
x+1<1+1
x+1<0

So, |x+1|=(x+1)
|x+1|=x1 .................... (1)

x<1
x1<11
x1<2

So, |x1|=(x1)
|x1|=x+1 ...................... (2)

Now, x,x1,x+1 are in A.P.

So,
2(x1)=x+(x+1)

2(x1)=1

x1=12

First term, x=121=32

Second term, |x+1|=32+1=12

Common difference, d=12(32)=2

Sum of n terms, Sn=n2{2a+(n1)d}

S20=10{2a+19d}

S20=10{2(32)+19×2}

S20=10(3+38)

S20=10(35)=350

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