Ratio and Proportion - IV

Question

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 $\frac{(a-d)^2(b^2-c^2)}{(bc)^2}$

Please help.....

May be this term : $(a-d)(\frac{1}{c^2}-\frac{1}{b^2})^2$ is $(a-d)(\frac{1}{c^2}-\frac{1}{b^2})$ not square of term

I assume that is supposed to be $(a-d)(\frac{1}{c^2}-\frac 1{b^2})^2$. Looks like you have an inappropriate ^ character... — Thomas Andrews, Mar 18 '13 at 14:59
For what it is worth, continued proportion means $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$. New term for me. — Thomas Andrews, Mar 18 '13 at 15:01
Yes, that is how we'd express it, but possibly a foreign term to OP. And I'm only guessing that since that was what came up when I googled. It doesn't seem to work for $(a,b,c,d)=(8,4,2,1)$, unless I botched my computation. — Thomas Andrews, Mar 18 '13 at 15:08
What if the $r<1$? The result might turn out the other way around? — Inceptio, Mar 18 '13 at 15:15
I think it is closer to true if $a,b,c,d$ are in arithmetic progression. It's still wrong, but it becomes much closer. — Thomas Andrews, Mar 18 '13 at 15:16
The meaning of continued proportion should be a defined properly. Only possibility is $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=1$, I suppose. I'm ain't very sure. — Inceptio, Mar 18 '13 at 15:20
If continued proportion is as Thomas Andrews says, a more common term would be geometric progression — Ross Millikan, Mar 18 '13 at 16:09

score 1 · Answer 1 · answered Mar 18 '13 at 15:26

If $a,b,c,d$ are in continued proportion, we can divide through by $a$ and make them $1,r,r^2,r^3$. Then: $$\begin {align} \left(\frac {a-b}c+\frac{a-c}b\right)^2-\left(\frac{d-b}c+\frac {d-c}b\right)^2&= \left(\frac {1-r}{r^2}+\frac{1-r^2}r\right)^2-\left( \frac {r^3-r}{r^2}+\frac{r^3-r^2}r\right)^2 \\&=\left( \frac{1-r^3}{r^2}\right)^2-\left(\frac{r^4-r}{r^2}\right)^2\\&=\frac{(1+r)^2(1-r^3)^2}{r^4}\\&=\frac{(r+r^2)^2(1-r^3)^2}{r^6}\\&=\frac{(b+c)^2(a-d)^2}{b^2c^2}\end {align}$$

Which differs from what you got for the right by a sign.

Hi, Thanks a lot that was great...thanks once again. – Sachin Sharmaa Mar 18 '13 at 15:57 — Sachin Sharmaa, Mar 18 '13 at 15:57

Ratio and Proportion - IV

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