Answer
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Hint: The given series is in the Geometrical Progression form as each consecutive term is multiplied by a fixed ratio. A Geometrical Progression is a sequence of numbers where each term is multiplied by its previous number of sequences with a constant number known as a common ratio\[r\]. In general, a common ratio \[r\] is found by dividing any term of the series with its previous term. The behavior of a geometric series depends on its common ratio.
If the ratio,
\[r = 1\]The progression is constant; all the terms in the series are the same.
\[r > 1\]The progression is increasing; all the subsequent terms in the series are increasing by the common factor.
\[r < 1\], the progression is decreasing; all the subsequent terms in the series are decreasing by the common factor.
Mathematically, a geometric progression series is summarized as \[{a_1},{a_1}r,{a_1}{r^2},{a_1}{r^3}........\]where \[{a_1}\] is the first term of series and $r$ is the common ratio.
The sum of a Geometrical decreasing series is given as\[{S_n} = \dfrac{a}{{1 - r}};r < 1\], whereas for a Geometrical increasing series \[{S_n} = \dfrac{a}{{r - 1}};r > 1\]
Complete step-by-step answer:
The series is in GP form with the first term\[a = \sqrt 2 \] and common ratio\[r = - \dfrac{1}{2}\] , where \[r < 1\] hence the series is geometrically decreasing.
Since a given series is a negative progression infinite series so, \[{S_n} = \dfrac{a}{{1 - r}};r < 1\] the formula is used.
Substitute \[a = \sqrt 2 \] and \[r = - \dfrac{1}{2}\] in the formula \[{S_n} = \dfrac{a}{{1 - r}};r < 1\] to determine the sum of the given GP series.
\[
{S_n} = \dfrac{a}{{1 - r}};r < 1 \\
= \dfrac{{\sqrt 2 }}{{1 - \left( { - \dfrac{1}{2}} \right)}} \\
= \dfrac{{\sqrt 2 }}{{\dfrac{3}{2}}} \\
= \dfrac{{2\sqrt 2 }}{3} \\
\]
Note: Before finding the sum of the given series, we need to find the nature of the series, i.e., AP, GP, or HP series. For GP series, the common ratio is a critical factor as it determines the nature of the series, whether increasing or decreasing.
If the ratio,
\[r = 1\]The progression is constant; all the terms in the series are the same.
\[r > 1\]The progression is increasing; all the subsequent terms in the series are increasing by the common factor.
\[r < 1\], the progression is decreasing; all the subsequent terms in the series are decreasing by the common factor.
Mathematically, a geometric progression series is summarized as \[{a_1},{a_1}r,{a_1}{r^2},{a_1}{r^3}........\]where \[{a_1}\] is the first term of series and $r$ is the common ratio.
The sum of a Geometrical decreasing series is given as\[{S_n} = \dfrac{a}{{1 - r}};r < 1\], whereas for a Geometrical increasing series \[{S_n} = \dfrac{a}{{r - 1}};r > 1\]
Complete step-by-step answer:
The series is in GP form with the first term\[a = \sqrt 2 \] and common ratio\[r = - \dfrac{1}{2}\] , where \[r < 1\] hence the series is geometrically decreasing.
Since a given series is a negative progression infinite series so, \[{S_n} = \dfrac{a}{{1 - r}};r < 1\] the formula is used.
Substitute \[a = \sqrt 2 \] and \[r = - \dfrac{1}{2}\] in the formula \[{S_n} = \dfrac{a}{{1 - r}};r < 1\] to determine the sum of the given GP series.
\[
{S_n} = \dfrac{a}{{1 - r}};r < 1 \\
= \dfrac{{\sqrt 2 }}{{1 - \left( { - \dfrac{1}{2}} \right)}} \\
= \dfrac{{\sqrt 2 }}{{\dfrac{3}{2}}} \\
= \dfrac{{2\sqrt 2 }}{3} \\
\]
Note: Before finding the sum of the given series, we need to find the nature of the series, i.e., AP, GP, or HP series. For GP series, the common ratio is a critical factor as it determines the nature of the series, whether increasing or decreasing.
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