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

Question:

If abc are in G.P., prove that:

(i) $a\left(b^{2}+c^{2}\right)=c\left(a^{2}+b^{2}\right)$

(ii) $a^{2} b^{2} c^{2}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)=a^{3}+b^{3}+c^{3}$

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

(iv) $\frac{1}{a^{2}-b^{2}}+\frac{1}{b^{2}}=\frac{1}{b^{2}-c^{2}}$

(v) (a + 2b + 2c) (a − 2b + 2c) = a2 + 4c2.

Solution:

a, b and c are in G.P.

$\therefore b^{2}=a c$

(i) $\mathrm{LHS}=a\left(b^{2}+c^{2}\right)$

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

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

$=c\left(a^{2}+b^{2}\right)=$ RHS

(ii) LHS $=a^{2} b^{2} c^{2}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)$

$=\frac{b^{2} c^{2}}{a}+\frac{a^{2} c^{2}}{b}+\frac{a^{2} b^{2}}{c}$

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

$=a^{3}+b^{3}+c^{3}=\mathrm{RHS}$

(iii) LHS $=\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}$

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

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

$=\frac{(a+b+c)^{2}}{(a+b+c)(a-b+c)} \quad\left[\because(\mathbf{a}+\mathbf{b}+\mathbf{c})(\mathbf{a}-\mathbf{b}+\mathbf{c})=\mathbf{a}^{2}-\mathbf{b}^{2}+\mathbf{c}^{2}+2 \mathbf{a c}\right]$

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

$($ iv $) \mathrm{LHS}=\frac{1}{a^{2}-b^{2}}+\frac{1}{b^{2}}$

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

$=\frac{a^{2}}{\left(a^{2} b^{2}-b^{4}\right)}$

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

$=\frac{1}{a c-c^{2}}$

$=\frac{1}{b^{2}-c^{2}}=\mathrm{RHS}$

(v) $\mathrm{LHS}=(a+2 b+2 c)(a-2 b+2 c)$

$=a^{2}-4 b^{2}+4 c^{2}+4 a c$

$=a^{2}-4 a c+4 c^{2}+4 a c$         [Using(1)]

$=a^{2}+4 c^{2}=\mathrm{RHS}$