[Maths]Topic: Ratio and ProportionHelp!!!If b is the mean proportion b

1.	[Maths]Topic: Ratio and ProportionHelp!!!If b is the mean proportion between a and c, Prove:Thanks in advance. No Spammimg :)
Answer» $\texttt{\textsf{\large{\underline{Solution!}}}}$ Given: → b is the mean proporrtion between a and C. To Prove: $\tt\implies\dfrac{a^{2}-b^{2}+c^{2}}{a^{-2}-b^{-2}+c^{-2}} = b^{4}$ As b is the mean proportion between a and c: $\tt\implies \dfrac{a}{b}=\dfrac{b}{c}$ $\tt\implies b^{2}=ac$ Taking LHS: $\tt=\dfrac{a^{2}-b^{2}+c^{2}}{a^{-2}-b^{-2}+c^{-2}}$ $\tt=\dfrac{a^{2}-b^{2}+c^{2}}{\dfrac{1}{a^{2}}-\dfrac{1}{b^{2}}+\dfrac{1}{c^{2}}}$ $\tt=\dfrac{a^{2}-b^{2}+c^{2}}{\dfrac{b^{2}c^{2}-a^{2}c^{2}+a^{2}b^{2}}{a^{2}b^{2}c^{2}}}$ $\tt=\dfrac{a^{2}b^{2}c^{2}(a^{2}-b^{2}+c^{2})}{b^{2}c^{2}-a^{2}c^{2}+a^{2}b^{2}}$ $\tt=\dfrac{(ac)^{2}b^{2}(a^{2}-b^{2}+c^{2})}{b^{2}c^{2}-(ac)^{2}+a^{2}b^{2}}$ PUT b² = ac: $\tt=\dfrac{b^{4}\cdot b^{2}(a^{2}-b^{2}+c^{2})}{b^{2}c^{2}-b^{4}+a^{2}b^{2}}$ $\tt=\dfrac{b^{4}\cdot b^{2}(a^{2}-b^{2}+c^{2})}{b^{2}(c^{2}-b^{2}+a^{2})}$ $\tt=\dfrac{b^{4}(a^{2}-b^{2}+c^{2})}{c^{2}-b^{2}+a^{2}}$ $\tt=b^{4}$ = RHS! Hence, Proved. $\texttt{\textsf{\large{\underline{Learn More!}}}}$ Properties of Ratio and Proportion. 1. Invertendo: If a : b : : c : d then b : a : : d : c 2. Alternendo: If a : b : : c : d then a : c : : b : d 3. COMPONENDO: If a : b : : c : d then (a + b) : b : : (c + d) : d 4. Dividendo: If a : b : : c : d then (a - b) : b : : (c - d) : d 5. Componendo and Dividendo: If a : b : : c : d then (a + b) : (a - b) : : (c + d) : (c - d) 6. Convertendo: If a : b : : c : d then a : (a - b) : : c : (c - d)