If $6 a^{3} b, 12 a^{2} b^{2}, k$ and $48 a b^{3}$ are in the proportion, then express $k$ in terms of $a$ and $b$

24 a b^{2}

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24 a^{2} b^{2}

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24 a^{2} b

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4 a^{2} b^{2}

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Solution

The correct option is A $24 a^{2} b^{2}$
In, $6 a^{3} b : 12 a^{2} b^{2} :: k : 48 a b^{3}$ , product of extremes $=$ product of means
$6 a^{3} b \times 48 a b^{3} = 12 a^{2} b^{2} \times k$
$k = 24 a^{2} b^{2}$

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If 6a3b,12a2b2,k and 48ab3 are in the proportion, then express k in terms of a and b

The correct option is A 24a2b2In, 6a3b:12a2b2::k:48ab3, product of extremes = product of means6a3b×48ab3=12a2b2×kk=24a2b2

If $6 a^{3} b, 12 a^{2} b^{2}, k$ and $48 a b^{3}$ are in the proportion, then express $k$ in terms of $a$ and $b$

The correct option is A $24 a^{2} b^{2}$
In, $6 a^{3} b : 12 a^{2} b^{2} :: k : 48 a b^{3}$ , product of extremes $=$ product of means
$6 a^{3} b \times 48 a b^{3} = 12 a^{2} b^{2} \times k$
$k = 24 a^{2} b^{2}$