It is given that p,q,r are in G.P.
∴ q2 + pr
Now, px2 + 2qx + r = 0
⇒ px2 + 2\(\sqrt{prx}\) + r = 0
⇒ \((\sqrt{px} +\sqrt{r})^2\) = 0
⇒ \((\sqrt{px} +\sqrt{r})\) = 0
⇒ x = \(-\sqrt{\frac{r}{p}}\)
It is given that the equations px2, + 2px + r = 0 and dx2 + 2ex + f − 0 have a common root and the equation px2, + 2qx + r = 0 has equal roots equal to −\(\sqrt{\frac{r}{p}}\)
∴ −\(\sqrt{\frac{r}{p}}\) is a root of the equation dx2 + 2ex + f = 0
⇒ \(d\frac{r}{p} − 2e\sqrt{\frac{r}{p}} + f\) = 0
⇒ \(\frac{d}{p} − 2e\sqrt{\frac{1}{pe}} + \frac{f}{r}\) [Dividing though out by r]
⇒ \(\frac{d}{p} − \frac{2e}{q} +\frac{f}{r}\) = 0 [∴ q2 = pr]
⇒ 2\(\frac{e}{q}\) = \(\frac{d}{p}\) + \(\frac{f}{r}\)
⇒ \(\frac{d}{p}\), \(\frac{e}{q}\), \(\frac{f}{r}\) are in A. P
Hence proved