Q.Which number should be added to each of 4, 6, and 10 so that the resulting sums are in continued proportion?

Let the number to be added be \(x\). \(\therefore (4 + x):(6 + x) = (6 + x):(10 + x)\) Or, \(\frac{4 + x}{6 + x} = \frac{6 + x}{10 + x}\) Or, \((6 + x)^2 = (4 + x)(10 + x)\) Or, \(36 + 12x + x^2 = 40 + 4x + 10x + x^2\) Or, \(36 + 12x = 40 + 14x\) Or, \(12x - 14x = 40 - 36\) Or, \(-2x = 4\) Or, \(x = -2\) \(\therefore\) If \(-2\) is added to each of 4, 6, and 10, the resulting numbers will be in continued proportion.
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