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

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

1. A

$c$

2. B

$\sqrt{de}$

3. C

$e$

4. D

$d$

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### Solution:

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

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

As , are in A.P.

$e=2d-c=2cr-c=c\left(2r-1\right)$

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

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

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

$\begin{array}{l}⇒\frac{2a-b}{ab}=\frac{1}{c}\\ \therefore \frac{a{b}^{2}}{\left(2a-b{\right)}^{2}}=\frac{a{b}^{2}}{\left(ab/c{\right)}^{2}}=\frac{{c}^{2}}{a}\\ =c\left(2r-1\right)=e\end{array}$  