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If $a,b,c$ are in A.P., $b,c,d$ are in G.P. and $\large\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P., then prove that $a,c,e$ are in G.P.$

1 Answer

Toolbox:
  • If $x,y,z$ are in A.P. then $2y=x+z$
  • If $x,y,z$ are in G.P., then $y^2=xz$
Given $a,b,c$ are in A.P. $b,c,d$ are in G.P. and $\large\frac{1}{c},\frac{1}{d},\frac{1}{e}$ are in A.P.
$\Rightarrow\: 2b=a+c$.......(i)
$c^2=bd.$.......(ii) and $\large\frac{2}{d}=\frac{1}{c}+\frac{1}{e}$..........(iii)
(iii) $\Rightarrow\:2ce=d(c+e)$
We have to prove that $a,c,e$ are in G.P.
$\Rightarrow\:$ We have to prove that $c^2=ae$
Step 2
From (i) and (ii) we get $c^2=\large\frac{a+c}{2}$$d$.....(iv)
From (iii) $d=\frac{2ce}{c+e}$
Substituting this value of $d$ in (iv) we get
$c^2=\large\frac{a+c}{2}$$\times \large\frac{2ce}{c+e}$
$\Rightarrow\:c^2(c+e)=(a+c)ce$
$\Rightarrow\:c(c+e)=e(a+c)$
$\Rightarrow\:c^2+ce=ae+ce$
$\Rightarrow\:c^2=ae$
$\Rightarrow\:a,c,e$ are in G.P.
hence proved
answered Apr 2, 2014 by rvidyagovindarajan_1
 

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