Let a, b, c, d and and e be real numbers such that a>b>0 and c>0. if 1a,1b,1c are in A.P., b,c,d are in G.P. and c, d, e are in A. P., then ab2(2a−b)2 is equal to

A
$c$
B
$\sqrt{d e}$
C
$e$
D
$d$

Solution:

Let $r$ be the common ratio of the G.P. $b, c, d$ , then

$b = \frac{c}{r}, c = c, d = c r$

As $c, d, e$ , are in A.P.

$e = 2 d - c = 2 c r - c = c (2 r - 1)$

Also $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.

$\Rightarrow \frac{1}{a} = \frac{2}{b} - \frac{1}{c} = \frac{2 r}{c} - \frac{1}{c} = \frac{2 r - 1}{c}$

Also $\frac{2}{b} - \frac{1}{a} = \frac{1}{c}$

$\begin{array}{l} \Rightarrow \frac{2 a - b}{a b} = \frac{1}{c} \\ ∴ \frac{a b^{2}}{(2 a - b)^{2}} = \frac{a b^{2}}{(a b / c)^{2}} = \frac{c^{2}}{a} \\ = c (2 r - 1) = e \end{array}$

Let $a, b, c, d$ and and $e$ be real numbers such that $a > b > 0$ and $c > 0 .$ if $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., $b, c, d$ are in G.P. and $c, d, e$ are in A. P., then $\frac{a b^{2}}{(2 a - b)^{2}}$ is equal to

Solution:

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