If one A M , A and two geometric means G1 and G2 inserted between any two positive numbers,show that - Maths - Sequences and Series

Question

If one A M , A and two geometric means
G1 and G2 inserted between
any two positive numbers,show that G1
2 / G2 + G22
/ G1 = 2A

Ankush Jain · Accepted Answer

Let b and c be two positive numbers Then, A = A M of b and c $\Rightarrow A = \frac{b + c}{2} \Rightarrow 2 A = b + c$ Again, G1 and G2 be two geometric means between b and c Then, b, G1, G2 , c are in G P with common ratio $= {(\frac{c}{b})}^{\frac{1}{2 + 1}} = {(\frac{c}{b})}^{\frac{1}{3}}$ Therefore, G1 = br $= b {(\frac{c}{b})}^{\frac{1}{3}}$ and G2 = br2 $= b {(\frac{c}{b})}^{\frac{2}{3}}$ Now, $\frac{{G_{1}}^{2}}{G_{2}} + \frac{{G_{2}}^{2}}{G_{1}} = \frac{{G_{1}}^{3} + {G_{2}}^{3}}{G_{1} G_{2}} = \frac{{(b (\frac{c}{b})^{\frac{1}{3}})}^{3} + {(b (\frac{c}{b})^{\frac{2}{3}})}^{3}}{(b (\frac{c}{b})^{\frac{1}{3}}) (b (\frac{c}{b})^{\frac{2}{3}})} = \frac{b^{3} (\frac{c}{b}) + b^{3} {(\frac{c}{b})}^{2}}{b^{2} ((\frac{c}{b})^{\frac{1}{3} + \frac{2}{3}})} = \frac{b^{3} (\frac{c}{b}) [1 + \frac{c}{b}]}{b^{2} ((\frac{c}{b})^{\frac{3}{3}})} = \frac{b^{3} (\frac{c}{b}) [\frac{b + c}{b}]}{b^{2} (\frac{c}{b})} = b (\frac{b + c}{b}) = b + c = 2 A$ [Hence proved]

If one A.M., A and two geometric means G1 and G2 inserted between any two positive numbers,show that G1 2 / G2 + G22 / G1 = 2A.

If one A.M., A and two geometric means G₁ and G₂ inserted between any two positive numbers,show that G₁ ^₂/ G₂ + G_2²/ G₁= 2A.