Question

Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion ​

Answer 1

Step-by-step explanation:

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Answer 2

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$\bold {Solution:}$

Let the number $x$ to be subtracted from each of the given numbers.

Then the numbers $\large(12-x),(16-x) and (21-x)$ are in continued proportion. ... (Say)

∴ $\Large \frac{12-x}{16-x}$$\frac{16-x}{21-x} = k$

∴ k=$\Large \frac{12-x-(16-x)}{16-x-(21-x)} =$$\Large \frac{12-x-16+x}{16-x-21-x}$

$\Large =\frac{-4}{4}$$\Large = \frac{-5}{5}$

∴ $\Large \frac{12-x}{16-x} \frac {4}{5}$ ∴ $5(12-x) = 4(16-x)$

∴ $60-5x=64-6x \:\:\:\:\:\:\:\:\:\:∴ -5x+4x=64-60$

∴ $-x = 4 \:\:\:\:\:\:\:\:\:∴ x = -4$

Ans: (-4) should be subtracted from each of the given numbers.