If a, b, c are in GP, prove that

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