Accorrding to Question

$ \Rightarrow \frac{{{S_5}}}{{S{'_5}}} = 49$

(here, ${S_5} = $ Sum of $5$ terms and ${S_5} = $ Sum of their reciprocals)

$ \Rightarrow \frac{{\frac{{a\left( {{r^5} – 1} \right)}}{{\left( {r – 1} \right)}}}}{{\frac{{{a^{ – 1}}\left( {{r^{ – 5}} – 1} \right)}}{{\left( {{r^{ – 1}} – 1} \right)}}}} = 49$

$ \Rightarrow \frac{{a\left( {{r^5} – 1} \right) \times \left( {{r^{ – 1}} – 1} \right)}}{{{a^{ – 1}}\left( {{r^{ – 5}} – 1} \right) \times \left( {r – 1} \right)}} = 49$

$\frac{{{a_2}\left( {1 – {r^5}} \right) \times \left( {1 – r} \right) \times {r^5}}}{{\left( {1 – {r^5}} \right) \times \left( {1 – r} \right) \times r}} = 49$

$ \Rightarrow {a^2}{r^4} = 49 \Rightarrow {a^2}{r^4} = {7^2}$

$ \Rightarrow \boxed{a{r^2} = 7}\,\,\,\,\,\,\,\,\,\,…\left( 1 \right)$

Also, given ${S_1} + {S_3} = 35$

$a + a{r^2} = 35\,\,\,\,\,\,…..\left( 2 \right)$

Now substituting the value of eq. $(1)$ in eq. $(2)$

$a + 7 = 35$

$\boxed{a = 28}$