(hipótese):
Seja a P.G.$(...\,,\,a_{p-1}\,,\,a_{p}\,,\,a_{p+1}\,,\,...)$ de razão igual a $\,q\,$
(tese):
Queremos demonstrar que $\phantom{X}(ap)^2\,=\,a_{p-1}\,\centerdot\,a_{p+1}\phantom{X}$
(DEMONTRAÇÃO):
$\,\left\{\begin{array}{rcr} a_{p-1}\,=\,a_1\,q^{p-2}\;& \\ a_p\,=\,a_1\,q^{p-1}\phantom{Xx}& \\ a_{p+1}\,=\,a_1\,q^{p}\phantom{Xx}& \end{array} \right.\;\Longrightarrow$ $\;a_{p+1}\,\centerdot \,a_{p-1}\,=\,a_1\,q^{p-2}\,\centerdot \,a_1\,q^{p}\,=$ $\,(a_1)^2\,\centerdot\,q^{p-2}\,\centerdot \,q^{p}\,=$ $\,(a_1)^2\,\centerdot\,q^{2p-2}\,=$ $\,(a_1)^2\,\centerdot\,q^{2(p-1)}\,=$ $\,=\,(a_1\,q^{p-1})^2\,=$ $\,(a_p)^2$

c.q.d.