If a, b and c are in continued proportion, then prove that :

Since, a, b and c are in continued proportion.

$$\begin{aligned}\therefore \frac{a}{b} = \frac{b}{c} \\[1em] \Rightarrow b^2 = ac.\end{aligned}$$

Substituting, b² = ac in L.H.S. of equation $\frac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \frac{a}{c}$, we get :

$$\begin{aligned}\Rightarrow \frac{3a^2 + 5ab + 7ac}{3ac + 5bc + 7c^2} \\[1em] \Rightarrow \frac{a(3a + 5b + 7c)}{c(3a + 5b + 7c)} \\[1em] \Rightarrow \frac{a}{c}.\end{aligned}$$

Hence, proved that $\frac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \frac{a}{c}$.