Suppose 12x,|x+1|,|x−1| are in A.P., then sum to 10 terms of the A.P. is

# Suppose $\frac{1}{2}x,|x+1|,|x-1|$ are in A.P., then sum to 10 terms of the A.P. is

1. A

54

2. B

36

3. C

28

4. D

none of these

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### Solution:

As $\frac{1}{2}x,|x+1|,|x-1|$ are in A.P.

$2|x+1|=\frac{1}{2}x+|x-1|$                         (1)

If then (1) give us

$-2\left(x+1\right)=\frac{1}{2}x-\left(x-1\right)$

$⇒x=-2$

In this case common difference is 2 and sum to 10 terms is

$\frac{10}{2}\left[2\left(-1\right)+\left(9\right)\left(2\right)\right]=85$

If  $-1\le x<1$

$2\left(x+1\right)=\frac{1}{2}x-\left(x-1\right)⇒x=-2/5$

In this case common difference is 4/5, and sum to 10 terms is

$\frac{10}{2}\left[2\left(-\frac{1}{5}\right)+\left(9\right)\left(\frac{4}{5}\right)\right]=34$

If $x\ge 1$,then (1) becomes

$2x+2=\frac{1}{2}x+x-1$

$⇒\frac{1}{2}x=-3$

Not possible as $x\ge 1$