If a, b and c are in continued proportion, then prove that :
3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2=ca.
Since, a, b and c are in continued proportion.
∴ab=bc⇒b2=ac.\therefore \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow b^2 = ac.∴ba=cb⇒b2=ac.
Substituting, b2 = ac in L.H.S. of equation 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2=ca, we get :
⇒3a2+5ab+7ac3ac+5bc+7c2⇒a(3a+5b+7c)c(3a+5b+7c)⇒ac.\Rightarrow \dfrac{3a^2 + 5ab + 7ac}{3ac + 5bc + 7c^2} \\[1em] \Rightarrow \dfrac{a(3a + 5b + 7c)}{c(3a + 5b + 7c)} \\[1em] \Rightarrow \dfrac{a}{c}.⇒3ac+5bc+7c23a2+5ab+7ac⇒c(3a+5b+7c)a(3a+5b+7c)⇒ca.
Hence, proved that 3a2+5ab+7b23b2+5bc+7c2=ac\dfrac{3a^2 + 5ab + 7b^2}{3b^2 + 5bc + 7c^2} = \dfrac{a}{c}3b2+5bc+7c23a2+5ab+7b2=ca.
