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

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- Jul 30th 2010, 11:11 AMStuck Manarithmetic sequence
I can't find a common difference. Is that the way to start?

- Jul 30th 2010, 11:16 AMundefined
You can assume without loss of generality that $\displaystyle a \le b \le c$.

Edit: Actually, easier to assume $\displaystyle \frac{1}{b+c} \le \frac{1}{c+a} \le \frac{1}{a+b}$ (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. - Jul 30th 2010, 11:51 AMStuck Man
It doesn't say either sequence is in ascending order.

- Jul 30th 2010, 11:52 AMSoroban
Hello, Stuck Man!

Quote:

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

. . are three consecutive terms of an arithmetic sequence.

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

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

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

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

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

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

. . . . .$\displaystyle (b-a)(b+a) \:=\:(c-b)(c+b)$

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

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

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

. . $\displaystyle b^2$ is theof $\displaystyle a^2$ and $\displaystyle c^2$.*mean*

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

- Jul 30th 2010, 01:39 PMArchie Meade
Also, if

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

then it's only necessary to prove

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

Starting from the original sequence...

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

Multiply both sides by (a+c)

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

Multiply both sides by (b+c)

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

Multiply both sides by (a+b)

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

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

$\displaystyle b^2-a^2=c^2-b^2$ - Jul 30th 2010, 01:50 PMundefined
- Jul 31st 2010, 03:16 AMStuck 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.

- Jul 31st 2010, 03:35 AMBacterius
Note you can swap $\displaystyle a$ and $\displaystyle c$ 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.