Question

If a, b, c, d are in G.P., prove that:

(i) $\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}$

(ii) (a + b + c + d)² = (a + b)² + 2 (b + c)² + (c + d)²

(iii) (b + c) (b + d) = (c + a) (c + d)

Answer 1

a, b, c and d are in G.P.

$$\begin{gathered} \therefore b^2=ac \\ bc=ad \\ {c}^2=bd \end{gathered}$$ .......(1)

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=\frac{ab-cd}{b^2-c^2} \\ =\frac{ab-cd}{ac-bd}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{\left(ab-cd\right)b}{\left(ac-bd\right)b} \\ =\frac{a{b}^2-bcd}{\left(ac-bd\right)b} \\ =\frac{a\left(ac\right)-c\left(c^2\right)}{\left(ac-bd\right)b}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{a^{2}c-c^3}{\left(ac-bd\right)b} \\ =\frac{c\left(a^2-c^2\right)}{\left(ac-bd\right)b} \\ =\frac{\left(a+c\right)\left(ac-c^2\right)}{\left(ac-bd\right)b} \\ =\frac{\left(a+c\right)\left(ac-bd\right)}{\left(ac-bd\right)b}\left[\mathrm{Using}\left(1\right)\right] \\ =\frac{\left(a+c\right)}{b}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}={\left(a+b+c+d\right)}^2 \\ ={\left(a+b\right)}^2+2\left(a+b\right)\left(c+d\right)+{\left(c+d\right)}^2 \\ ={\left(a+b\right)}^2+2\left(ac+ad+bc+bd\right)+{\left(c+d\right)}^2 \\ ={\left(a+b\right)}^2+2\left(b^2+bc+bc+c^2\right)+{\left(c+d\right)}^2\left[\mathrm{Using}\left(1\right)\right] \\ ={\left(a+b\right)}^2+2{\left(b+c\right)}^2+{\left(c+d\right)}^2=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}=\left(b+c\right)\left(b+d\right) \\ =b^2+bd+bc+cd \\ =ac+c^2+ad+cd\left[\mathrm{Using}\left(1\right)\right] \\ =c\left(a+c\right)+d\left(a+c\right) \\ =\left(c+a\right)\left(c+d\right)=\mathrm{RHS} \end{gathered}$$