### What is the value of $n$ if the sum of $n$ terms of the $G.P. \space 2, \dfrac{ 2 }{ 3 }, \dfrac{ 2 }{ 9 } \space ... is \dfrac{ 80 }{ 27 }$?

$4$
1. In the given $G.P.,$ the first term, $a = 2$ and
the common ratio, $r = \dfrac{ a_{k+1} }{ a_k } \text{ where } k ≥ 1$
$\implies r = \dfrac{ \dfrac{ 2 }{ 3 } } { 2 } = \dfrac{ 1 }{ 3 }$
2. The sum of first $n$ terms of this $G.P.$ is given by,
\begin{align} S_n = & \dfrac{ a(1 - r^n) }{ (1 - r) } = \dfrac{ 80 }{ 27 } \\ \implies & \dfrac { 2 \left(1 - \left( \dfrac{ 1 }{ 3 } \right)^n \right) } { 1 - \dfrac{ 1 }{ 3 } } = \dfrac{ 80 }{ 27 } \\ \implies & \dfrac { 2 \left(1 - \left( \dfrac{ 1 }{ 3 } \right)^n \right) } { \dfrac{ 2 }{ 3 } } = \dfrac{ 80 }{ 27 } \\ \implies & \dfrac{ 6 }{ 2 } \left[ 1 - \left( \dfrac{ 1 }{ 3 } \right)^n \right] = \dfrac{ 80 }{ 27 } \\ \implies & \left[ 1 - \left( \dfrac{ 1 }{ 3 } \right)^n \right] = \dfrac{ 80 }{ 81 } \\ \implies & \left( \dfrac{ 1 }{ 3 } \right)^n = 1 - \dfrac{ 80 }{ 81 } \\ \implies & \dfrac{ 1 } { 3^n } = \dfrac{ 1 } { 81 } \\ \implies & 3^n = 3 ^ 4 \\ \implies & n = 4 \end{align}