Therefore ${A_1} = \frac{{a + {A_2}}}{2},\;{A_2} = \frac{{{A_1} + b}}{2}$
$ \Rightarrow $ ${A_1}+{A_2} = \frac{{a +b+{A_1}+{A_2}}}{2}$
$ \Rightarrow $ $ \frac{{{A_1}+{A_2}}}{2}$ $=\frac{{{a+b}}}{2}$ or ${A_1} + {A_2} = a + b$ …..$(i)$
and $a,\;{G_1},\;{G_2},\;b$ are in $G.P.$
Therefore $G_1^2 = a{G_2},\;G_2^2 = b{G_1}$ …..$(ii)$
$ \Rightarrow G_1^2G_2^2 = ab{G_1}{G_2}$
$\Rightarrow {G_1}{G_2} = ab$
Hence $\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}} = \frac{{a + b}}{{ab}}$
Trick : Let $a = 1,\;b = 2$, then ${A_1} + {A_2} = 1 + 2 = 3$
and ${G_1}\;.\;{G_2} = 2 \times 1 = 2$
$\therefore \;$$\frac{{{A_1} + {A_2}}}{{{G_1}{G_2}}} = \frac{3}{2}$,
which is given by $(a).$
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$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$. If the composition of $f, \underbrace{(f \circ f \circ f \circ \ldots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to....................