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Question

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

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Solution

## $a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}\therefore 2b=a+c.......\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Also},b,c\mathrm{and}d\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}\therefore {c}^{2}=bd.......\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{And}\frac{1}{c},\frac{1}{d}\mathrm{and}\frac{1}{e}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}\therefore \frac{2}{d}=\frac{1}{c}+\frac{1}{e}\phantom{\rule{0ex}{0ex}}⇒d=\frac{2ce}{c+e}.......\left(\mathrm{iii}\right)\phantom{\rule{0ex}{0ex}}\because {c}^{2}=bd\left[\mathrm{From}\left(\mathrm{ii}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{c}^{2}=\left(\frac{a+c}{2}\right)\left(\frac{2ce}{c+e}\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{iii}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{c}^{2}\left(c+e\right)=ce\left(a+c\right)\phantom{\rule{0ex}{0ex}}⇒{c}^{2}+ce=ae+ec\phantom{\rule{0ex}{0ex}}⇒{c}^{2}=ae\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},a,c\mathrm{and}e\mathrm{are}\mathrm{also}\mathrm{in}\mathrm{G}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}$

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