Sum 3 times greater than $n^\text{th}$ term?

Relevant wiki: Geometric Progression Sum

Let the first term and common ratio of the geometric progression be $a = 36$ and $r = \frac 43$ respectively. Then, we have:

$\begin{aligned} \sum_{k=1}^n ar^k & > 3ar^{n-1} \\ \frac {a(r^n-1)}{r-1} & > 3ar^{n-1} \\ r^n-1 & > 3r^{n-1}(r-1) \\ r^{n+1}-1 & > 3r^{n-1}\left(\frac 43-1\right) \\ r^n-1 & > r^{n-1} \\ r^n - r^{n-1} & > 1 \\ r^n \left(1 - \frac 1r\right) & > 1 \\ r^n \left(1 - \frac 34 \right) & > 1 \\ r^n \left(\frac 14 \right) & > 1 \\ \left(\frac 43 \right)^n & > 4 \\ n & > \frac {\log 4}{\log \frac 43} \approx 4.819 \\ \implies n & = \boxed{5} \end{aligned}$

Let $a$ be the first term and $r$ be the common ratio

$a=36$ $r=\frac { 4 }{ 3 }$

$3({ ar }^{ n-1 }) < \frac { a(1-{ r }^{ n }) }{ 1-r }$

$3{ \times 36(\frac { 4 }{ 3 }) }^{ n-1 }<\frac { 36(1-{ \frac { 4 }{ 3 } }^{ n }) }{ 1-\frac { 4 }{ 3 } } \\$

$108\left( { \frac { 4 }{ 3 } } \right) ^{ n-1 }<\frac { 36(1-{ \frac { 4 }{ 3 } }^{ n }) }{ \frac { -1 }{ 3 } }$

$108\left( { \frac { 4 }{ 3 } } \right) ^{ n-1 }<-108(1-{ \frac { 4 }{ 3 } }^{ n })$

$81\left( { \frac { 4 }{ 3 } } \right) ^{ n }<-108(1-{ \frac { 4 }{ 3 } }^{ n })$

$81\left( { \frac { 4 }{ 3 } } \right) ^{ n }<-108+108\left( { \frac { 4 }{ 3 } }^{ n } \right) \\$

$-27\left( { \frac { 4 }{ 3 } } \right) ^{ n }<-108$

$27\left( { \frac { 4 }{ 3 } } \right) ^{ n }>108$

$\left( { \frac { 4 }{ 3 } } \right) ^{ n }>4\\$

$\log _{ \frac { 4 }{ 3 } }{ \left( { \frac { 4 }{ 3 } }^{ n } \right) >\log _{ \frac { 4 }{ 3 } }{ \left( 4 \right) } }$

$n>4.8188..$

$\implies n= \boxed{5}$