Let $a_n$ be the $n^{th}$ term of a G.P. of positive terms. If $\sum _{n=1}^{100}{a}_{2n+1}=200$  and $\sum _{n=1}^{100}{a}_{2n}$$\sum _{n=1}^{100}{a}_{2n}=100,$ then $\sum _{n=1}^{200}{a}_n$ is  equal to

$a_3+a_5+\mathrm{.......}+a_{201}=\frac{a_3[{\left(r^2\right)}^{100}-1]}{(r^2-1)}=200$  and  $a_2+a_4+\mathrm{.......}+a_{200}=\frac{a_2[{\left(r^2\right)}^{100}-1]}{(r^2-1)}=100$
Division gives $\frac{a_3}{a_2}=r=2$ . Hence  $a_2=\frac{100\left(3\right)}{4^{100}-1}\Rightarrow a_1=\frac{150}{4^{100}-1}$