For a sequence and . Then [IMAGE] is

JEE MainMathematicsModerateSeries and progressions

For a sequence $\{ a_{n}\},a_{1} = 2$ and
$\frac{a_{n + 1}}{a_{n}} = \frac{1}{3}$ . Then [IMAGE] is

Show answer and solution

a_{1} = 2;

\frac{a_{n + 1}}{a_{n}} = 1/3

\frac{a_{2}}{a_{1}} = 1/3 \Rightarrow a_{2} = 2/3

\frac{a_{3}}{a_{2}} = 1/3 \Rightarrow a_{3} = \frac{2}{3} \cdot \frac{1}{3}

\Rightarrow a_{4} = \frac{2}{3} \cdot \left( \frac{1}{3} \right)^{2}

\Rightarrow a_{r} = \frac{2}{3}\left( \frac{1}{3} \right)^{r - 2}

Now,

S = \sum_{\gamma = 1}^{20}\mspace{2mu} a_{r} = a_{1} + a_{2} + - \cdots + a_{20}

= 2 + \frac{2}{3} + \frac{2}{3}\left( \frac{1}{3} \right) + \cdots + \frac{2}{3}\left( \frac{1}{3} \right)^{18}

= 2 + \frac{2}{3}\left\lbrack 1 + \frac{1}{3} + \left( \frac{1}{3} \right)^{2} + + \left( \frac{1}{3} \right)^{18} \right\rbrack

= 2 + \frac{2}{3}\left\lbrack \frac{1 \cdot \left( 1 - \left( \frac{1}{3} \right)^{19} \right)}{1 - 1/3} \right\rbrack

= 2 + \frac{2}{3}\left\lbrack \frac{1 - \left( \frac{1}{3} \right)19}{\frac{2}{3}} \right\rbrack

= 2 + 1 - \left( \frac{1}{3} \right)^{19}

= 3\left( 1 - \frac{1}{3}20 \right)

Source: Grass-G / M-06 Mathematics Paper 14 October.docx

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