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Exercise 5.4 · Q8

Q.If ax=by=cza^x = b^y = c^z such that a,ba, b and cc are in GP and x,yx, y and zz are unequal positive integers, then show that 2y=1x+1z\dfrac{2}{y} = \dfrac{1}{x}+\dfrac{1}{z}.

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Introduce the common value k=ax=by=czk=a^x=b^y=c^z, express a,b,ca,b,c as powers of kk, then apply the GP condition b2=acb^2=ac and equate exponents.

If p,q,rp,q,r are in GP then q2=prq^2=pr (the square of the middle term equals the product of the outer terms). Also, if km=knk^{m}=k^{n} with k>0, k≠1k>0,\ k\neq1, then m=nm=n.

  1. Let ax=by=cz=ka^x=b^y=c^z=k (a common positive constant, since x,y,zx,y,z are positive integers and a,b,ca,b,c are terms of a GP, so positive).
  2. Solve each equation for aa, bb, cc in terms of kk:

a=k1/x,b=k1/y,c=k1/za=k^{1/x},\qquad b=k^{1/y},\qquad c=k^{1/z}

  1. Given a,b,ca,b,c are in GP, the middle term squared equals the product of the outer terms:

b2=acb^2=ac

  1. Substitute the expressions from Step 2:

(k1/y)2=k1/x⋅k1/z\left(k^{1/y}\right)^2=k^{1/x}\cdot k^{1/z}

  1. Simplify both sides using laws of exponents: …

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