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Question

If a1/x=b1/y=c1/z and a , b , c be in G . P . then x , y , z, in H . P .

A
True
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B
False
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Solution

The correct option is B False
Let
a1/x=b1/y=c1/z=k

So,
a=kx,b=ky,c=kz

Since,
a,b,c are in G.P

So,
b2=ac

Now,
k2y=kx.kz
k2y=kx+z

Therefore,
2y=x+z

Hence, x,y,z are in A.P

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