If $a, b, c$ are in GP, prove that $\left(a^{2}+b^{2}\right),(a b+b c),\left(b^{2}+c^{2}\right)$ are in GP.
To prove: $\left(a^{2}+b^{2}\right),(a b+b c),\left(b^{2}+c^{2}\right)$ are in GP
Given: $a, b, c$ are in GP
Formula used: When $a, b, c$ are in GP, $b^{2}=a c$
Proof: When a,b,c are in GP,
$b^{2}=a c \ldots(i)$
Considering $\left(a^{2}+b^{2}\right),(a b+b c),\left(b^{2}+c^{2}\right)$
$(a b+b c)^{2}=\left(a^{2} b^{2}+2 a b^{2} c+b^{2} c^{2}\right)$
$=\left(a^{2} b^{2}+a b^{2} c+a b^{2} c+b^{2} c^{2}\right)$
$=\left(a^{2} b^{2}+b^{4}+a^{2} c^{2}+b^{2} c^{2}\right)$ [From eqn. (i)]
$=\left[b^{2}\left(a^{2}+b^{2}\right)+c^{2}\left(a^{2}+b^{2}\right)\right]$
$(a b+b c)^{2}=\left[\left(b^{2}+c^{2}\right)\left(a^{2}+b^{2}\right)\right]$
From the above equation we can say that $\left(a^{2}+b^{2}\right),(a b+b c),\left(b^{2}+c^{2}\right)$ are in GP
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