If a- 4- bare in AP and a- 2- bare in GP then 1-a- 1- 1-b are in

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Selecting Consecutive Terms in GP

If a, 4, bare...

Question

If $a, 4, b$ are in $A P$ and $a, 2, b$ are in $G P$ then $\frac{1}{a}, 1, \frac{1}{b}$ are in

$A P$

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$G P$

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$H P$

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none of these

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Solution

The correct option is A
$A P$

Explanation for the correct option:
Step 1. Evaluating the given conditions:
$∵ a, 4, b$ are in $A P$
$\Rightarrow 8 = a + b$ …(1) $[∵ 2 b = a + b]$
$∵ a, 2, b$ are in $G P$
$\Rightarrow 4 = a b$ …(2) $[∵ b^{2} = a b]$
Step 2. Divide equation (1) by (2), we get
$\Rightarrow$ $\frac{(a + b)}{a b} = \frac{8}{4}$
$\Rightarrow \frac{1}{b} + \frac{1}{a} = 2$
$∴ \frac{1}{a}, 1, \frac{1}{b}$ are in $A P$
Hence, option A. is the correct answer:

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If a,4,bare in AP and a,2,bare in GP then 1a,1,1b are in

If $a, 4, b$ are in $A P$ and $a, 2, b$ are in $G P$ then $\frac{1}{a}, 1, \frac{1}{b}$ are in