Hi Bunuel,

I took the LCM of b=1+1/4a

Then b became

5/4aI got the answer as -85/256

Why can't we take the LCM and do it this way?

And if we can then either way, we must get the same answer..

Please help...thanks[/quote]

Hi

zanaik89,

Refer the highlighted portion, It should be :

\(b=1+\frac{1}{4}a\)-------------(2)

=\(\frac{4+a}{4}\) (Though we don't require this step)

a=1+\(\frac{1}{4} + \frac{1}{16} + \frac{1}{64}\)=\(\frac{64+16+4+1}{64}\)=\(\frac{85}{64}\)------------(1)

Hence \(a-b=\frac{85}{64}-(1+\frac{1}{4}*\frac{85}{64})=\frac{85}{64}-1-\frac{85}{4*64}=\frac{(4*85)-(4*64)-85}{4*64}\)=-\(\frac{1}{256}\)

Another Method:-\(a=1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)

So, \(\frac{1}{4}a=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}\) (Dividing both sides by 4)

So, \(b=1+\frac{1}{4}a=1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256}\)

Hence, \(a-b=(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64})-(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\frac{1}{256})\)

Or, \(a−b=−\frac{1}{256}\)

_________________

Regards,

PKN

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