# Ratio and Proportion II

• Mar 18th 2013, 06:56 AM
sachinrajsharma
Ratio and Proportion II
If a,b,c,d are continued proportion : Prove that : $(\frac{a-b}{c}+\frac{a-c}{b})^2-(\frac{d-b}{c}+\frac{d-c}{b})^2=(a-d)^(\frac{1}{c^2}-\frac{1}{b^2})^2$

After solving LH.S I got : $\frac{2(a-d)}{(bc)^2}$

But after solving R.H.S I am getting $\frac{(a-d)^2(b^2-c^2)}{(bc)^2}$

• Mar 18th 2013, 08:52 PM
Punch
Re: Ratio and Proportion II
notice that it is of the form $A^2-B^2$, you should try manipulating using the following formula, $A^2-B^2=(A+B)(A-B)$
• Mar 18th 2013, 08:58 PM
sachinrajsharma
Re: Ratio and Proportion II
If u noticed that I have already done..... if u can do it further than please let me know..
• Mar 19th 2013, 05:56 AM
wholesalejerseys
Re: Ratio and Proportion II
I have for k=1,2,3... where is a term of some infinite sequence { }

Some manipulation:

Is there anything I can do with this now?

Is it legal to multiply the sums like this?

The second factor looks like a geometric series and I should be able to get the sum of the terms equal to some constant A (if series converges).

(for each k)

Does that seem
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• Mar 19th 2013, 09:24 PM
ibdutt
Re: Ratio and Proportion II
Please recheck there is something amiss. Notice the powers of variables on both the sides??