Question

$\bold\pink{Question:-}$
Which term of the following sequences:
$(a) \: 2 ,2 \sqrt{2} ,4,...is \: 128 \: ?$
$(b) \: \sqrt{3} ,3,3 \sqrt{3} ,...is \: 729?$
$(c) \: \frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,...is \: \frac{1}{19683} ?$
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Answer 1

$\huge\bold\pink{Answer:-}$

$\bold{The \: Given \: sentences \: is \: 2 ,2 \sqrt{2} ,4,...}$

$\bold\pink{Here, }a = 2 \: and \: r \: = \: \frac{2 \sqrt{2} }{2} = \sqrt{2}$

Let the${n}^{th}$ term of the given sequence be 127.

$= > \bold{(2)( \sqrt{2} {)}^{n - 1} = 128 }$

$= > \bold{ (2)(2 {)}^{ \frac{n - 1}{2} } = {(2)}^{7} }$

$= > \bold{w {(2)}^{ \frac{n - 1}{2} } { + }^{1} = {(2)}^{7} }$

$\bold{∴ \: \frac{n - 1}{2} + 1 = 7 }$

$= > \bold{ \frac{n - 1}{2} + 1 = 7 }$

$= > \bold{n - 1 = 12}$

$= > \bold{n = 13}$

$\bold\pink{Thus,}$

The ${13}^{th}$ term of the given sequence is 128.

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$the \: given \: sequence \: is \sqrt{3},3 ,3 \sqrt{3} ,...$

$a \: = \sqrt{3} \: and \: r \: = \frac{3}{ \sqrt{3} } = \sqrt{3}$

Let the ${n}^{th}$ term of the given sequence be 729.

$\bold{an \: = {ar}^{n - 1} }$

$\bold{∴ \: {ar}^{n - 1} = 723 }$

$= > \bold{( \sqrt{ 3} )( \sqrt{3} {)}^{ n - 1} = 729 }$

$\bold{ = > (3) \frac{1}{2} (3) \frac{n - 1}{2} = {(3)}^{6} }$

$\bold{ = > (3 {)}^{ \frac{1}{2} + \frac{n - 1}{2} } = (3 {)}^{6} }$

$\bold{∴ \: \frac{1}{2} + \frac{n - 1}{2} = 6}$

$\bold{ = > \frac{1 + n - 1}{2} = 6 }$

$\bold{n = 12}$

$\bold\pink{Thus,}$

The ${12}^{th}$ term of the given sequence is 728

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$the \: given \: sequence \: is \: \frac{1}{3} , \frac{1}{9} , \frac{1}{27} , ...$

$\bold\pink{Here,}a = \frac{1}{3 } \: and \: r \: = \frac{1}{9} + \frac{1}{3} = \frac{1}{3}$

Let the ${n}^{th}$ term of the given sequence be $\frac{1}{19683}$

$\bold{an \: = {ar}^{n - 1} }$

$\bold{∴a {r}^{n - 1} }$

$\bold{ = > ( \frac{1}{3} ) (\frac{1}{3} {)}^{n - 1} = \frac{1}{19683} }$

$\bold{ {( \frac{1}{3} )}^{n}( \frac{1}{3} {)}^{9} }$

$\bold{n \: = 9}$

$\bold\pink{Thus,}$

The ${9}^{th}$ term of the given sequence is $\frac{1}{19683}$

Answer 2

Answer:

Refers to the attachment....