# A problem on progression

If a,b,c are in arithmetic progression., p,q,r in harmonic progression and ap,bq,cr are in geometric progression., then $\frac{p}{r}+\frac{r}{p} =$ ?

EDIT: I have tried to use the basic/standard properties of the respective progressions to get the desired result, but I am not yet successful.

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Let's represent your arithmetic progression as

$$a,a+h,a+2h$$

where $h=b-a$

and your harmonic progression as

$$p,\frac{p}{l+1},\frac{p}{2l+1},$$

(following Wikipedia)

and the condition that the products of corresponding terms of the arithmetic and harmonic sequences form a geometric sequence implies

$$\frac{a+h}{a(l+1)}=\frac{a+2h}{a+h}\frac{l+1}{2l+1}$$

Since

$$\frac{p}{r}+\frac{r}{p}=2l+1+\frac1{2l+1}$$

we solve for $l$, which has two solutions.

I leave the rest to you. ;)

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Notice that $\rm\:\ \quad\displaystyle \frac{p}r+\frac{r}p\ =\ \frac{(p+r)^2}{pr} - 2$

But we have that$\rm\quad\displaystyle p\:r\ =\ \frac{(bq)^2}{ac}\ \$ via $\rm\ ap,\:bq,\:cr\$ geometric

and we have $\rm\quad\ \ \displaystyle p+r\ =\ \frac{2pr}q\quad\ \ \$ via $\rm\ p,q,r\$ harmonic

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If a,b,c are in arithmetic progression, then you can write them as a, a+d, a+2d. Maybe all you need is 2b=a+c. Similarly for geometric progression, you can write them as ap, apz, apz^2 where z is the common ration (often we use r, but that was taken). Again, maybe (bq)^2=ap*cr is all you need.

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I knew this relations and I tried to use them but not quite successful. – Damir Frezivioski Oct 27 '10 at 15:54