If a, b, c are in G.P., prove that the following are also in G.P.:

Solution:

a, b and c are in G.P.

$\therefore b^{2}=a c \quad \ldots \ldots(1)$

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

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

Therefore, $a^{2}, b^{2}$ and $c^{2}$ are also in $\mathrm{G} . \mathrm{P} .$

(ii) $\left(b^{3}\right)^{2}=\left(b^{2}\right)^{3}=(a c)^{3} \quad[$ Using $(1)]$

$\Rightarrow\left(b^{3}\right)^{2}=a^{3} c^{3}$

Therefore, $a^{3}, b^{3}$ and $c^{3}$ are also in G.P.

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

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

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

 

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

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

Therefore, $\left(a^{2}+b^{2}\right),\left(b^{2}+c^{2}\right)$ and $(a b+b c)$ are also in G. P.