Progression (Sequence and Series)HardBloom L3
Question
If $A_{1},{\text{ }A}_{2}$ be $A.M$.'s, $G_{1},G_{2}$ be $G.M$.'s and $H_{1},H_{2}$ be H.M.'s between two numbers, then
Options
A.$\frac{G_{1}G_{2}}{H_{1}H_{2}} = \frac{A_{1} + A_{2}}{H_{1} + H_{2}}$
B.$\frac{G_{1}G_{2}}{I_{1}I_{2}} = \frac{H_{1} + H_{2}}{\Lambda_{1} + \Lambda_{2}}$
C.$\frac{G_{1}G_{2}}{{\text{ }A}_{1}{\text{ }A}_{2}} = \frac{H_{1} + H_{2}}{{\text{ }A}_{1} + A_{2}}$
D.$\frac{G_{1}G_{2}}{\Lambda_{1}\Lambda_{2}} = \frac{A_{1} + A_{2}}{I_{1} + H_{2}}$
Solution
Sol. [Key Concept: For two A.M.s, G.M.s, H.M.s between two numbers: $\frac{G_{1}G_{2}}{H_{1}H_{2}} = \frac{A_{1 +}A_{2}}{H_{1 +}H_{2}}$.]
$$A_{1} + A_{2} = a + b;G_{1}G_{2} = ab$$
$$\begin{matrix} & & \frac{1}{H_{1}} + \frac{1}{H_{2}} = \frac{1}{a} + \frac{1}{\text{ }b} & = \frac{a + b}{ab} \\ \Rightarrow & & & \frac{H_{1} + H_{2}}{H_{1}H_{2}} = \frac{A_{1} + A_{2}}{G_{1}G_{2}} \end{matrix}$$
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