Carcass wrote:

If \(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) , then \(x =\)

A. \(3\)

B. \(1\)

C. \(\frac{1}{3}\)

D. \(-\frac{1}{3}\)

E. \(-3\)

\(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\) .....

We can see the formation of the terms is similar on both sides, so let us get the left hand side in terms similar to RHS, by taking out \(\frac{x}{3}\)..

\(\frac{x}{3} - \frac{x}{6} + \frac{x}{9} - \frac{x}{12} =\frac{x}{3}*1 - \frac{x}{3}*\frac{1}{2} + \frac{x}{3}*\frac{1}{3} - \frac{x}{3}*\frac{1}{4} = \frac{x}{3}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4})\)

This should be equal to RHS..

\(\frac{x}{3}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}........\frac{x}{3}=1....x=3\) .....

A

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Some useful Theory.

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