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

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

$\therefore b^{2}=a c$   ...(i)

$b c=a d$

$c^{2}=b d$

(i) $\mathrm{LHS}=\frac{a b-c d}{b^{2}-c^{2}}$

$=\frac{a b-c d}{a c-b d} \quad[\operatorname{Using}(1)]$

$=\frac{(a b-c d) b}{(a c-b d) b}$

$=\frac{a b^{2}-b c d}{(a c-b d) b}$

$=\frac{a(a c)-c\left(c^{2}\right)}{(a c-b d) b} \quad[\operatorname{Using}(1)]$

$=\frac{a^{2} c-c^{3}}{(a c-b d) b}$

$=\frac{c\left(a^{2}-c^{2}\right)}{(a c-b d) b}$

$=\frac{a^{2} c-c^{3}}{(a c-b d) b}$

$=\frac{c\left(a^{2}-c^{2}\right)}{(a c-b d) b}$

$=\frac{(a+c)\left(a c-c^{2}\right)}{(a c-b d) b}$

$=\frac{(a+c)(a c-b d)}{(a c-b d) b} \quad[\mathrm{U} \sin \mathrm{g}(1)]$

$=\frac{(a+c)}{b}=$ RHS

(ii) LHS $=(a+b+c+d)^{2}$

$=(a+b)^{2}+2(a+b)(c+d)+(c+d)^{2}$

$=(a+b)^{2}+2(a c+a d+b c+b d)+(c+d)^{2}$

$=(a+b)^{2}+2\left(b^{2}+b c+b c+c^{2}\right)+(c+d)^{2} \quad$ [Using (1)]

$=(a+b)^{2}+2(b+c)^{2}+(c+d)^{2}=\mathrm{RHS}$

(iii) LHS $=(b+c)(b+d)$

$=b^{2}+b d+b c+c d$

$=a c+c^{2}+a d+c d$         [Using (1)]

$=c(a+c)+d(a+c)$

$=(c+a)(c+d)=\mathrm{RHS}$