arithmetic sequence

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
I can't find a common difference. Is that the way to start?
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.
 
Last edited:
Oct 2009
461
5
It doesn't say either sequence is in ascending order.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, Stuck Man!

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 the mean of \(\displaystyle a^2\) and \(\displaystyle c^2\).


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

 
Dec 2009
3,120
1,342
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)

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

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

\(\displaystyle b^2-a^2=c^2-b^2\)
 
Last edited:
Oct 2009
461
5
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.
 
Nov 2009
927
260
Wellington
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 \(\displaystyle a\) 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.