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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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