Question

given that a^x,b^y,c^z,d^u are in gp then x,y,z,u are in ​

Answer 1

Answer:

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Step-by-step explanation:

Given that, ax=by=cz=du.

Suppose that, ax=by=cz=du=k.

∴a=k1x,b=k1y,c=k1z,and,d=k1u.........(1).

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

∴ba=cb=dc.

By (1), then, k1yk1x=k1zk1y=k1uk1z,or,

k1y−1x=k1z−1y=k1u−1z,

⇒(1y−1x)=(1z−1y)=(1u−1z),

⇒1x,1y,1z,1u, are in A.P.

⇒x,y,z,u are in H.P.