# Progression Prolem............

• Nov 3rd 2010, 09:13 PM
Arka
Progression Prolem............
If one Arithmetic Mean (A) and two Geometric Means (p,q) are inserted between two given numbers, show that

$\displaystyle {(p^2)/q}+{(q^2)/p}=2A$
• Nov 4th 2010, 03:52 AM
HallsofIvy
Call the original numbers x and y. There arithmetic mean is $\displaystyle A= \frac{x+ y}{2}$. Strictly speaking, two numbers have only one geometric mean but I think that you mean that x, p, q, y form a geometric progression. That would mean that p= xr, [tex]q= xr^2[tex] and $\displaystyle y=xr^3$. Dividing y by x, $\displaystyle \frac{y}{x}= r^3$ so that $\displaystyle r= \sqrt[3]{\frac{x}{y}}$ and so $\displaystyle p= x(\frac{x^{1/3}}{y^{1/3}})= \frac{x^{4/3}}{y^{1/3}}$ and $\displaystyle q= x\frac{x^{2/3}}{y^{2/3}}= \frac{x^{5/3}}{y^{2/3}}$.

Put those into $\displaystyle \frac{p^2}{q}+ \frac{q^2}{p}$ and see if you get 2A= x+ y.
• Nov 4th 2010, 09:51 AM
Arka
No I'm not getting x+y , if you are getting it then please show me the calculation......