1. If a,b,c are in G.P and ax=by=cz ,then x,y,z are in a)A.P b)G.P c)H.P

Dear swapnil,
a,b,c are in GP then y²= ac
Also a^x=b^y=c^z
On solving we get y=2xz/(x+z)
which means x,y,z are in H.P (C) is correct option
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Approved

a,b,c are in GP

b²=ac,    a^x=b^y=c^z=k

              a=k^1/x,b=k^1/y,c=k^1/z

so b²=ac

  k^2/y₌k^1/x+1/y

2/y=i/x+1/z

so x,y,z are in HP

If a,b,c are consecutive terms of a.p , a^2, b^2, c^2 are consecutive terms of g.p., a+b+c=3/4 find a