In an increasing, geometric series,

Solution:

$a r+a r^{5}=\frac{25}{2}$

$a r^{2} \times a r^{4}=25$

$a^{2} r^{6}=25$

$a r^{3}=5$

$a=\frac{5}{r^{3}}$

$\frac{5 r}{r^{3}}+\frac{5 r^{5}}{r^{3}}=\frac{25}{2}$

$\frac{1}{r^{2}}+r^{2}=\frac{5}{2}$

Put $r^{2}=t$

$\frac{t^{2}+1}{t}=\frac{5}{2}$

$2 t^{2}-5 t+2=0$

$2 t^{2}-4 t-t+2=0$

$(2 t-1)(t-2)=0$

$t=\frac{1}{2}, 2 \Rightarrow r^{2}=\frac{1}{2}, 2$

$r=\sqrt{2}$

$=a r^{3}+a r^{5}+a r^{7}$

$=a r^{3}\left(1+r^{2}+r^{4}\right)$

$=5[1+2+4]=35$