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in Geometric Progressions by (15.7k points)

If a, b, c are in GP and a1/x = b1/y = c1/z then prove that x, y, z are in AP.

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It is given that: 

a 1/x = b1/y = c1/z 

Let a1/x = b1/y = c1/z = k 

⇒ a 1/x = k 

⇒ (a1/x) x = kx…(Taking power of x on both sides.) 

⇒ a 1/x × x = k

⇒ a = kx 

Similarly b = ky 

And c = kz 

It is given that a,b,c are in G.P. 

⇒ b2 = ac 

Substituting values of a,b,c calculated above, we get: 

⇒ (ky )2 = kxk z 

⇒ k2y = kx + z 

Comparing the powers we get, 

2y = x + z 

Which is the required condition for x,y,z to be in A.P. 

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

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