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Given that a, b, c and d are in geometric progression prove that:

$(b-c)^2 + (c-a)^2 + (d-b)^2 = (a-d)^2$

I've established that ar = b, br = c, cr = d where r is a common ratio.

However I do not understand what my next approach should be.

Thanks

I've done this, thus far:

$(ar-ar^2)^2+(ar^2-a)^2+(ar^3-ar)^2$

Sort of unrelated to this, but is there a quick way to expand expressions like these?

I found this: $a^2r^6-2a^2r^3+a^2$

But how can I divide by $a^2$ if this an expression and not an equation?

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1  
Try expressing everything in terms of $a,r$. Substitute. Cancel out $a$. Square and establish an identity with what remains. –  Macavity Aug 17 '13 at 14:11
    
Hint 2: The equation you have needs to be simplified (expressed in terms of only two variables, say $a, r$. Use $b = ar, c = ar^2, ...$ –  Macavity Aug 17 '13 at 14:26
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2 Answers

up vote 0 down vote accepted

Set $b = ar$, $c=ar^2$ and $d=ar^3$.

Then the left side of your equation is

\begin{align} &(ar-ar^2)^2 + (ar^2-a)^2 + (ar^3-ar)^2 \\ &= a^2r^2(1-r)^2+a^2(r^2-1)^2+a^2r^2(r^2-1)^2\\ &= a^2(1-r)^2(r^2+(r+1)^2+r^2(r+1)^2)\\ &=a^2(1-r)^2(r^2+r+1)^2\\ &=(a(1-r^3))^2=(a-d)^2\end{align}

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did you factor $a^2$? I factor $a^2$ and get $a^2(r^2(1-r)^2+(r^2-1)^2+r^2(r^2-1)^2)$ –  salman Aug 17 '13 at 14:55
    
@user90771 Yes. In the third line, I factored $a^2$ and also a $(1-r)^2$. –  Alraxite Aug 17 '13 at 14:59
    
How can you factor the $(1-r)^2$? How does $\frac{a^2(r^2-1)^2}{a^2(1-r)^2} = (r+1)^2$ –  salman Aug 17 '13 at 15:06
    
@user90771 Yes. Use $r^2-1 = (r-1)(r+1)$ to see this. –  Alraxite Aug 17 '13 at 15:10
    
sorry I still didn't understand. $\frac{((r-1)(r+1))^2}{(1-r)^2}$ = ? (how does this simplify into $(r+1)^2$? Is $(r-1) = (1-r)$? If so how/why? –  salman Aug 17 '13 at 15:13
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Clearly, we need to eliminate $b,c$ from the left hand side

Assuimng $a\cdot b\cdot c\cdot d\ne0,$

$$\text{ As }\frac ba=\frac cb=\frac dc,\text{ we have } b^2=ca,c^2=bd, ad=bc$$

$$(b-c)^2 + (c-a)^2 + (d-b)^2 $$

$$= b^2+c^2-2bc+c^2+a^2-2ca+d^2-2bd+b^2$$

$$=a^2+d^2+2(b^2+c^2-bc-ca-bd)$$

$$=a^2+d^2+2(ca+bd-ad-ca-bd)$$

$$=a^2+d^2-2ad=(a-d)^2$$

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fantastic method thanks –  salman Aug 17 '13 at 15:46
    
@user90771, my pleasure. Sometimes looking at the Right Hand side helps optimizing the answer –  lab bhattacharjee Aug 17 '13 at 15:48
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