If the geometric series has first term [tex]a[/tex] and common ratio [tex]r[/tex], then its [tex]N[/tex]-th partial sum is
[tex]\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}[/tex]
Multiply both sides by [tex]r[/tex], then subtract [tex]rS_N[/tex] from [tex]S_N[/tex] to eliminate all the middle terms and solve for [tex]S_N[/tex] :
[tex]rS_N = ar + ar^2 + ar^3 + \cdots + ar^N[/tex]
[tex]\implies (1 - r) S_N = a - ar^N[/tex]
[tex]\implies S_N = \dfrac{a(1-r^N)}{1-r}[/tex]
The [tex]N[/tex]-th partial sum for the series of reciprocal terms (denoted by [tex]S'_N[/tex]) can be computed similarly:
[tex]\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}[/tex]
[tex]\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}[/tex]
[tex]\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}[/tex]
[tex]\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}[/tex]
We're given that [tex]a=1[/tex], and the sum of the first [tex]n[/tex] terms of the series is
[tex]S_n = \dfrac{1-r^n}{1-r} = 364[/tex]
and the sum of their reciprocals is
[tex]S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}[/tex]
By substitution,
[tex]\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243[/tex]
Manipulating the [tex]S_n[/tex] equation gives
[tex]\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363[/tex]
so that substituting again yields
[tex]r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}[/tex]
and it follows that
[tex]r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}[/tex]
