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Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$

if $\frac{1}{M}:G$ is 4:5 then a:b can be:

Option 1)
$1:4$

Option 2)
$1:2$

Option 3)
$2:3$

Option 4)
$3:4$

Answers (2)

best_answer

As we learnt in

Arithmetic mean of two numbers (AM) -

$A=\frac{a+b}{2}$

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

and

Geometric mean of two numbers (GM) -

$GM= \sqrt{ab}$

- wherein

It is to be noted that a,G,b are in GP and a,b are two non - zero numbers.

Given G²= ab

$2M=\frac{1}{a}+\frac{1}{b}$

and $\frac{\frac{1}{M}}{G}=\frac{4}{5}$

$\therefore GM=\frac{5}{4}$

$\therefore G^{2}M^{2}=\frac{25}{16}$

$=ab\times \left (\frac{a+b}{2ab} \right )^{2}= \frac{25}{16}$

$=\frac{ab\times \left ( a+b \right )^{2}}{4a^{2}b^{2}}= \frac{25}{16}$

$\therefore 4a^{2}+4b^{2}-17ab=0$

$\therefore 4\left ( \frac{a}{b} \right )^{2}-17\left ( \frac{a}{b} \right )+4=0$

$\therefore \frac{a}{b}=1:4$

Option 1)

$1:4$

Option 2)

$1:2$

Option 3)

$2:3$

Option 4)

$3:4$

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