If x- y and z are in A P- then x- y-1 and z are in-A- APB- GPC- HPD- a or bE- none of these

The correct option is C HPOption (c) Take y=2 and x+y/1 xy = 1 and (y+z)/1 yz= 3 When y=2. x= 1/3 and z = 1/7, 2( 1/3)(1/7)/(1/7) (1/3)=1/2=y 1 ... h ...

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Quantitative Aptitude

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If x, y and z...

Question

If x,y and z are in AP, then x, $y^{- 1}$ and z are in?

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a or b

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none of these

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Solution

The correct option is C HP
Option (c)
Take y=2 and $\frac{x + y}{1 - x y}$ = 1 and $\frac{(y + z)}{1 - y z}$ = 3

When y=2. x= $\frac{- 1}{3}$ and z = $\frac{1}{7}$ ,

$\frac{2 (\frac{- 1}{3}) (\frac{1}{7})}{(\frac{1}{7}) - (\frac{1}{3})} = \frac{1}{2} = y - 1 . . .$ hence they are in HP

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If x,y and z are in AP, then x, y−1 and z are in?

The correct option is C HPOption (c) Take y=2 and x+y1−xy = 1 and (y+z)1−yz= 3 When y=2. x=−13 and z = 17, 2(−13)(17)(17)−(13)=12=y−1... hence they are in HP

If x,y and z are in AP, then x, $y^{- 1}$ and z are in?

The correct option is C HP
Option (c)
Take y=2 and $\frac{x + y}{1 - x y}$ = 1 and $\frac{(y + z)}{1 - y z}$ = 3

When y=2. x= $\frac{- 1}{3}$ and z = $\frac{1}{7}$ ,

$\frac{2 (\frac{- 1}{3}) (\frac{1}{7})}{(\frac{1}{7}) - (\frac{1}{3})} = \frac{1}{2} = y - 1 . . .$ hence they are in HP