If a, b, c are in A.P., b,c,d are in G.P. and

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:

$a, b$ and $c$ are in A.P.

$\therefore 2 b=a+c \quad \ldots \ldots .(\mathrm{i})$

Also, $b, c$ and $d$ are in G.P.

$\therefore c^{2}=b d \quad \ldots \ldots$ (ii)

And $\frac{1}{c}, \frac{1}{d}$ and $\frac{1}{e}$ are in A.P.

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

$\Rightarrow d=\frac{2 c e}{c+e} \quad \ldots \ldots$ (iii)

$\because c^{2}=b d \quad[$ From (ii) $]$

$\Rightarrow c^{2}=\left(\frac{a+c}{2}\right)\left(\frac{2 c e}{c+e}\right) \quad[$ Using $($ i $)$ and $($ iii $)]$

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

$\Rightarrow c^{2}+c e=a e+e c$

$\Rightarrow c^{2}=a e$

Therefore, $a, c$ and $e$ are also in G.P.

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