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8. Sequences and Series
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જો $\text{y}\,=\,{{\text{x}}^{\frac{\text{1}}{\text{3}}}}\text{.}\,{{\text{x}}^{\frac{\text{1}}{\text{9}}}}\text{.}\,{{\text{x}}^{\frac{\text{1}}{\text{27}}}}\,.....\,\infty  $ હોય, તો $\text{y}\,=......$

A

$x^{1/3}$

B

$x^{2/3}$

C

$x^{1/2}$

D

$x$

Solution

$y\,\,=\,\,{{x}^{\frac{1}{3}}}\,.\,\,{{x}^{\frac{1}{9}}}\,.\,{{x}^{\frac{1}{27}}}\,\,……..\,\,\infty $

$y\,\,=\,\,{{x}^{\frac{1}{3}\,+\,\frac{1}{9}\,+\,\frac{1}{27}}}\,……..\,\,\infty $

$y\,\,=\,\,\,{{x}^{\left( \frac{a}{1\,-\,r} \right)}}\,=\,\,{{x}^{\left( \frac{{}^{1}\!\!\diagup\!\!{}_{3}\;}{1\,-\,{}^{1}\!\!\diagup\!\!{}_{3}\;} \right)}}\,\,=\,\,{{x}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$

$y\,\,=\,\,{{x}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$

Standard 11
Mathematics

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