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Selecting Consecutive Terms in GP

Between two u...

Question

Between two unequal numbers, if $a_{2}, a_{2}$ are two AMs; $g_{1}, g_{2}$ are two GMs and $h_{1}, h_{2}$ are two HMs then $g_{1} . g_{2}$ is equal to-

a_{1} h_{1}

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a_{1} h_{2}

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a_{2} h_{2}

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a_{2} h_{1}

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Solution

The correct options are
C $a_{1} h_{2}$
D $a_{2} h_{1}$
Le A, $a_{1}, a_{2}$ , B be in A.P.
$∴ a_{1} = A + \frac{B - A}{3} = \frac{2 A + B}{3}$
$∴ a_{2} = A + 2. \frac{B - A}{3} = \frac{A + 2 B}{3}$
Also $A, g_{1}, g_{2}, B$ are in G.P.
$∴ \frac{B}{A} = r^{3}$
$∴ g_{1} = A r = A (B / A)^{1 / 3}$
$∴ g_{2} = A r^{2} = A (B / A)^{2 / 3}$
$∴ g_{1} g_{2} = A^{2} (B / A) = A B$
Also A, $h_{1}, h_{2}, B$ are in H.P.
$∴ \frac{1}{A}, \frac{1}{h_{1}}, \frac{1}{h_{2}}, \frac{1}{B}$ are in AP
$∴ \frac{1}{h_{1}} = \frac{1}{A} + \frac{1}{3} (\frac{1}{B} - \frac{1}{A}) = \frac{1}{A} + \frac{A - B}{3 A B}$
$\Rightarrow \frac{1}{h_{1}} = \frac{3 B + A - B}{3 A B} = \frac{A + 2 B}{3 A B} \Rightarrow h_{1} = \frac{3 A B}{A + 2 B}$ and
$\frac{1}{h_{2}} = \frac{1}{A} + \frac{2}{3} (\frac{1}{B} - \frac{1}{A}) = \frac{3 B + 2 (A - B)}{3 A B}$
$\Rightarrow \frac{1}{h_{2}} = \frac{2 A + B}{3 A B}$
$∴ h_{2} = \frac{3 A B}{2 A + B}$
Obviously $g_{1} g_{2} = A B = a_{1} h_{2} = a_{2} h_{1}$
Hence, option 'B' and 'D' are correct.

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A_{1}, A_{2}, G_{1}, G_{2} .

Two AMs $A_{1}$ and $A_{2}$ , two GMs $G_{1}$ and $G_{2}$ and two HMs $H_{1}$ and $H_{2}$ are inserted between two numbers a and b, then

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