arithmetic sequence

**Stuck Man** · July 30th 2010, 11:11 AM

I can't find a common difference. Is that the way to start?

**undefined** · July 30th 2010, 11:16 AM

Originally Posted by Stuck Man

I can't find a common difference. Is that the way to start?

You can assume without loss of generality that $a \le b \le c$ .

Edit: Actually, easier to assume $\frac{1}{b+c} \le \frac{1}{c+a} \le \frac{1}{a+b}$ (perhaps this is what they meant anyway), and then equate (the second term minus the first term) with (the third term minus the second term) which will naturally lead to the desired result.

**Stuck Man** · July 30th 2010, 11:51 AM

It doesn't say either sequence is in ascending order.

**Soroban** · July 30th 2010, 11:52 AM

Hello, Stuck Man!

8. It is given that: . $\dfrac{1}{b+c},\;\dfrac{1}{a+c},\;\dfrac{1}{a+b}$

. . are three consecutive terms of an arithmetic sequence.

Show that $a^2,\:b^2,\:c^2$ are also three consecutive terms on an arithmetic sequence.

The difference of consecutive terms is the common difference, $d.$

$\dfrac{1}{a+c} - \dfrac{1}{b+c} \:=\:d \quad\Rightarrow\quad b-a \:=\:d(a+c)(b+c) \;\;[1]$

$\dfrac{1}{a+b} - \dfrac{1}{a+c} \:=\:d \quad\Rightarrow\quad c-b \:=\:d(a+b)(a+c)\;\;[2]$

Divide [1] by [2]: . $\dfrac{b-a}{c-b} \;=\;\dfrac{d(a+c)(b+c)}{d(a+b)(a+c)}$

and we have: . $\dfrac{b-a}{c-b} \:=\:\dfrac{b+c}{a+b}$

. . . . . $(b-a)(b+a) \:=\<img src=$ c-b)(c+b)" alt="(b-a)(b+a) \:=\

c-b)(c+b)" />

. . . . . . . . . $b^2-a^2 \:=\:c^2 - b^2$

. . . . . . . . . . . . $2b^2 \:=\:a^2+c^2$

. . . . . . . . . . . . $b^2 \;=\;\dfrac{a^2+c^2}{2}$

. . $b^2$ is the mean of $a^2$ and $c^2$ .

Therefore, $a^2,\:b^2,\:c^2$ form an arithmetic sequence.

**Archie Meade** · July 30th 2010, 01:39 PM

Also, if

$a^2,\ b^2,\ c^2$ are consecutive terms in an arithmetic series,

then it's only necessary to prove

$b^2-a^2=c^2-b^2=common\ difference$

Starting from the original sequence...

$\frac{1}{a+c}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{a+c}$

Multiply both sides by (a+c)

$1-\frac{a+c}{b+c}=\frac{a+c}{a+b}-1$

Multiply both sides by (b+c)

$b+c-(a+c)=\frac{(b+c)(a+c)}{a+b}-(b+c)$

Multiply both sides by (a+b)

$(b-a)(a+b)=(b+c)(a+c)-(b+c)(a+b)=(b+c)(a+c-a-b)=(b+c)(c-b)$

$ba+b^2-a^2-ab=bc-b^2+c^2-cb$

$b^2-a^2=c^2-b^2$

**undefined** · July 30th 2010, 01:50 PM

Originally Posted by Stuck Man

It doesn't say either sequence is in ascending order.

Without loss of generality - Wikipedia, the free encyclopedia

**Stuck Man** · July 31st 2010, 03:16 AM

I still don't know why undefined is saying that a <= b <= c. An arithmetic sequence can have a negative common difference and then a >= b >= c.

**Bacterius** · July 31st 2010, 03:35 AM

Originally Posted by Stuck Man

I still don't know why undefined is saying that a <= b <= c. An arithmetic sequence can have a negative common difference and then a >= b >= c.

Note you can swap $a$ and $c$ and still get the same result. This is what Undefined meant by "without loss of generality" ... the result Undefined gave can be adapted in trivial ways to any possible context, although it would be too long (and boring) to enumerate them all, so we just say WLOG.

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