Prove that

To prove: p, q, r are in GP

Given: $\left(p^{2}+q^{2}\right),(p q+q r),\left(q^{2}+r^{2}\right)$ are in GP

Formula used: When $a, b, c$ are in GP, $b^{2}=a c$

Proof: When $\left(p^{2}+q^{2}\right),(p q+q r),\left(q^{2}+r^{2}\right)$ are in GP,

$(p q+q r)^{2}=\left(p^{2}+q^{2}\right)\left(q^{2}+r^{2}\right)$

$p^{2} q^{2}+2 p q^{2} r+q^{2} r^{2}=p^{2} q^{2}+p^{2} r^{2}+q^{4}+q^{2} r^{2}$

$2 p q^{2} r=p^{2} r^{2}+q^{4}$

$p q^{2} r+p q^{2} r=p^{2} r^{2}+q^{4}$

$p q^{2} r-q^{4}=p^{2} r^{2}-p q^{2} r$

$q^{2}\left(p r-q^{2}\right)=p r\left(p r-q^{2}\right)$

$q^{2}=p r$

From the above equation we can say that p, q and r are in G.P.