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 } ^@?


Answer:

^@ 4 ^@

Step by Step Explanation:
  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} @^

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