Question

If a , b, c are in AP ; b , c, d are in G. P and 1/c, 1/d, 1/e are in AP. Prove that a, c, e are on G. P.

How to solve after this so that
$c {}^{2} = ae$

Answer 1

Here is your answer..........
Thanks

Answer 2

$i) a , b \: and \: c \: are \: in \: A.P :$

$b = \frac{a+c}{2} \: ---(1)$

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

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

$\implies \frac{2}{d} = \frac{e+c}{ce}$

$\implies \frac{d}{2} = \frac{ce}{c+e}$

$\implies d = \frac{2ce}{c+e}\: ---(2)$

$iii) b,c \: and \: d \: are \: in \: G.P$

$c^{2} = bd\: ---(3)$

$\implies c^{2} = \frac{(a+c)}{2} \times \frac{2ce}{(c+e)}\:[From\:(1) \:and \:(2)]$

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

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

$\implies c^{3} + c^{2}e = ace + c^{2}e$

$\implies c^{3} = ace$

$\implies c^{2} = ae$

$\therefore \green { a,c \:and \: e} \: are \:in \:G.P$

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