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Which term of the G.P. (i)

    \[\mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2},\text{ }\mathbf{4},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{128}?\]

(ii)

    \[\mathbf{1},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{1}/\mathbf{9},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{1}/\mathbf{243}\]

Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Which term of the G.P. (i)

    \[\mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2},\text{ }\mathbf{4},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{128}?\]

(ii)

    \[\mathbf{1},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{1}/\mathbf{9},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{1}/\mathbf{243}\]

From the question it is given that,

Last term = 128

First term a = 2,

Then,

    \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( 2\surd 2 \right)\text{ }\div \text{ }\left( 2 \right) \\ r\text{ }=\text{ }\left( 2\surd 2 \right)/2 \\ r\text{ }=\text{ }\surd 2 \\ \end{array}\]

Then,

    \[\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}} \\ So,\text{ }128\text{ }=\text{ }2{{\left( \surd 2 \right)}^{n\text{ }\text{ }1}} \\ {{2}^{7}}~=\text{ }2{{\left( \surd 2 \right)}^{n\text{ }\text{ }1}} \\ {{2}^{7}}/2\text{ }=\text{ }{{\left( \surd 2 \right)}^{n\text{ }\text{ }1}} \\ {{2}^{7\text{ }\text{ }1}}~=\text{ }{{\left( \surd 2 \right)}^{n\text{ }\text{ }1}} \\ {{2}^{6}}~=\text{ }{{\left( \surd 2 \right)}^{n\text{ }\text{ }1}} \\ {{\left( \surd 2 \right)}^{n\text{ }-1}}~=\text{ }{{\left( \surd 2 \right)}^{12}} \\ \end{array}\]

Now, comparing the powers

    \[\begin{array}{*{35}{l}} n\text{ }\text{ }1\text{ }=\text{ }12 \\ n\text{ }=\text{ }12\text{ }+\text{ }1 \\ n\text{ }=\text{ }13 \\ \end{array}\]

Therefore, 128 is the 13th term.

(ii)

    \[\mathbf{1},\text{ }\mathbf{1}/\mathbf{3},\text{ }\mathbf{1}/\mathbf{9},\text{ }\ldots \text{ }\mathbf{is}\text{ }\mathbf{1}/\mathbf{243}\]

Solution:-

From the question it is given that,

Last term (an) = 1/243

First term a = 1,

Then

    \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( 1/3 \right)\text{ }\div \text{ }\left( 1 \right) \\ r\text{ }=\text{ }\left( 1/3 \right)\text{ }\times \text{ }\left( 1/1 \right) \\ r\text{ }=\text{ }1/3 \\ \end{array}\]

Then,

    \[\begin{array}{*{35}{l}} {{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}} \\ 1/243\text{ }=\text{ }1\text{ }\times \text{ }{{\left( 1/3 \right)}^{n\text{ }\text{ }1}} \\ {{\left( 1/3 \right)}^{5}}~=\text{ }{{\left( 1/3 \right)}^{n\text{ }\text{ }1}} \\ \end{array}\]

By comparing both left hand side and right hand side,

    \[\begin{array}{*{35}{l}} 5\text{ }=\text{ }n\text{ }\text{ }1 \\ n\text{ }=\text{ }5\text{ }+\text{ }1 \\ n\text{ }=\text{ }6 \\ \end{array}\]

Therefore,

    \[1/243\text{ }is\text{ }{{6}^{th}}~\]

term.

i

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