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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.
Solution
\[\text { a, b and c are in A . P }. \]
\[ \therefore 2b = a + c . . . . . . . (i)\]
\[\text { Also, b, c and d are in G . P } . \]
\[ \therefore c^2 = bd . . . . . . . (ii)\]
\[\text {And } \frac{1}{c}, \frac{1}{d} \text { and } \frac{1}{e} \text { are in A . P .} \]
\[ \therefore \frac{2}{d} = \frac{1}{c} + \frac{1}{e} \]
\[ \Rightarrow d = \frac{2ce}{c + e} . . . . . . . (iii)\]
\[ \because c^2 = bd \left[ \text { From }(ii) \right] \]
\[ \Rightarrow c^2 = \left( \frac{a + c}{2} \right)\left( \frac{2ce}{c + e} \right) \left[ \text { Using } (i) \text { and } (iii) \right]\]
\[ \Rightarrow c^2 \left( c + e \right) = ce\left( a + c \right)\]
\[ \Rightarrow c^2 + ce = ae + ec\]
\[ \Rightarrow c^2 = ae\]
\[\text { Therefore, a, c and e are also in G . P } . \]
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