Let the fractions inserted b/w \(\frac{1}{4} & \frac{1}{2}\) be \(\frac{1}{x} and \frac{1}{y}\)We can easily calculate the value of \(\frac{1}{x}\) by:-

\(2(\frac{1}{x}-\frac{1}{4})=\frac{1}{2}-\frac{1}{x}\)Solving we get, \(\frac{1}{x}=\frac{1}{3}\)Also, \(\frac{2}{y}=\frac{1}{3}+\frac{1}{2}\)

Solving we get \(\frac{1}{y}=\frac{5}{12}\)Hence, the 4 fractions are \(\frac{1}{4}, \frac{1}{3}, \frac{5}{12} and \frac{1}{2}\)

The required sum is thus:-

\(sum=\frac{1}{4}+ \frac{1}{3}+ \frac{5}{12}+ \frac{1}{2}=\frac{18}{12}\)

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