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Question

If a+bxabx=b+cxbcx=c+dxcdx(x0), then show that a, b, c and d are in G.P.

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Solution

Given :

a+bxabx=b+cxbcx=c+dxcdx

Now, a+bxabx=b+cxbcx

Applying componendo and dividendo

(a+bx)+(abx)(a+bx)(abx)=(b+cx)+(bcx)(b+cx)(bcx)

2a2bx=2b2cx

ab=bc

Similiarly,

((b+cx)+(bcx)(b+cx)(bcx))=((c+dx)+(cdx)(c+dx)(cdx))

bc=cd.

Therefore, a, b, c and d are in G.P.


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