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# If a, b, c are in A.P.; b, c, d are in G.P. and 1 by c 1 by d 1 by e are in A.P. prove that a, c, e are in G.P.

20.   If a, b, c are in A.P.; b, c, d are in G.P. and 1/c , 1/d , 1/e are in A.P. prove that a, c, e are in G.P.

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Given: a, b, c are in A.P

$b-a=c-b..............................1$

Also,  b, c, d are in G.P.

$c^2=bd..............................2$

Also, 1/c, 1/d, 1/e are in A.P

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}...........................3$

To prove: a, c, e are in G.P. i.e.$c^2=ae$

From 1, we get  $2b=a+c$

$b=\frac{a+c}{2}$

From 2, we get

$d=\frac{c^2}{b}$

Putting values of b and d, we get

$\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}$

$\frac{2}{d}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2b}{c^2}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{2(a+c)}{2c^2}=\frac{1}{c}+\frac{1}{e}$

$\Rightarrow \frac{(a+c)}{c^2}=\frac{e+c}{ce}$

$\Rightarrow \frac{(a+c)}{c}=\frac{e+c}{e}$

$\Rightarrow e(a+c)=c(e+c)$

$\Rightarrow ea+ec=ec+c^2$

$\Rightarrow ea=c^2$

Thus, a, c, e are in G.P.

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