If $$Sn=1+13+132+$$..$$+13n$$−$$1$$,n∈N, then the least value of ' n ' such that $$3$$−$$2Sn$$<$$1100$$ is

$$Sn=1+13+132+$$..$$+13n$$−$$1$$⇒$$Sn=11-13n1-13$$ sum of n terms in G.$$P=$$ $$a1-rn1-r3$$−$$2Sn$$ $$100$$⇒n−$$1$$≥$$5$$ ⇒n≥$$6$$

Geometric progression

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$If S_{n} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . + \frac{1}{3^{n - 1}}, n \in N, then the least value of ' n' such that 3 - 2 S_{n} < \frac{1}{100} is$

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Solution

$S_{n} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . + \frac{1}{3^{n - 1}}$
$\Rightarrow S_{n} = \frac{1 (1 - \frac{1}{3^{n}})}{1 - \frac{1}{3}}$ $(s u m o f n t e r m s i n G . P = \frac{a (1 - r^{n})}{1 - r})$
$3 - 2 S_{n} < \frac{1}{100}$
$\Rightarrow$ $3 - \frac{2 (1 - \frac{1}{3^{n}})}{1 - 1 / 3} < \frac{1}{100} \Rightarrow \frac{1}{3^{n - 1}} < \frac{1}{100}$
$\Rightarrow 3^{n-1} > 100 \Rightarrow n - 1 \geq 5 \Rightarrow n \geq 6$

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Geometric progression

Remember concepts with our Masterclasses.

If Sn=1+13+132+..+13n−1,n∈N, then the least value of ' n ' such that 3−2Sn<1100 is

Sn=1+13+132+..+13n−1⇒Sn=11-13n1-13 sum of n terms in G.P= a1-rn1-r3−2Sn<1100 ⇒3−21−13n1−1/3<1100⇒13n−1<1100⇒3n-1>100⇒n−1≥5 ⇒n≥6

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$If S_{n} = 1 + \frac{1}{3} + \frac{1}{3^{2}} + . . + \frac{1}{3^{n - 1}}, n \in N, then the least value of ' n' such that 3 - 2 S_{n} < \frac{1}{100} is$