Question 1014418
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if a,b,c are in an arithmetic progression and x,y,z are in a geometric progression prove that x^b*y^c*z^a = x^c*y^a*z^b 
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<pre>
Since a, b, c are in an arithmetic progression, we have 
   a = b - d, c = b + d.

Since x, y, z are in an geometric progression, we have 
   x = y*r^(-1), z = y*r.

Now,   x^b*y^c*z^a = {{{(y*r^(-1))^b}}} . {{{y^(b+d)}}} . {{{(y*r)^(b-d)}}} = {{{y^(b+(b+d)+b-d)}}} . {{{r^(-b + (b-d))}}} = {{{y^(3b)}}}.{{{r^(-d)}}}.

Next,  x^c*y^a*z^b = {{{(y*r^(-1))^(b+d)}}} . {{{y^(b-d)}}} . {{{(y*r)^b)}}} = {{{y^((b+d)+b-d + b)}}} . {{{r^(-(b+d)+b)}}} = {{{y^(3b)}}}.{{{r^(-d)}}}.

Right sides are the same.
So are left sides.

Proved.
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