# Thread: Ratio and Proportion II

1. ## Ratio and Proportion II

If a,b,c,d are continued proportion : Prove that : $\displaystyle (\frac{a-b}{c}+\frac{a-c}{b})^2-(\frac{d-b}{c}+\frac{d-c}{b})^2=(a-d)^(\frac{1}{c^2}-\frac{1}{b^2})^2$

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

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

2. ## Re: Ratio and Proportion II

notice that it is of the form $\displaystyle A^2-B^2$, you should try manipulating using the following formula, $\displaystyle A^2-B^2=(A+B)(A-B)$

3. ## Re: Ratio and Proportion II

If u noticed that I have already done..... if u can do it further than please let me know..

4. ## 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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5. ## Re: Ratio and Proportion II

Please recheck there is something amiss. Notice the powers of variables on both the sides??