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Determine the

    \[\mathbf{1}{{\mathbf{2}}^{\mathbf{th}}}~\]

term of a G.P. whose

    \[{{8}^{\mathbf{th}}}~\]

term is

    \[192\]

and common ratio is

    \[2\]

.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
Determine the

    \[\mathbf{1}{{\mathbf{2}}^{\mathbf{th}}}~\]

term of a G.P. whose

    \[{{8}^{\mathbf{th}}}~\]

term is

    \[192\]

and common ratio is

    \[2\]

.

From the question it is given that,

a8 =

    \[192\]

and r =

    \[2\]

Then, by the formula

    \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}} {{a}_{8}}~=\text{ }a{{r}^{8\text{ }\text{ }1}} \\ 192\text{ }=\text{ }a{{\left( 2 \right)}^{8\text{ }\text{ }1}} \\ 192\text{ }=\text{ }a{{\left( 2 \right)}^{7}} \\ a\text{ }=\text{ }192/{{2}^{7}} \\ a\text{ }=\text{ }192/128 \\ a\text{ }=\text{ }3/2 \\ \end{array}\]

Now,

    \[\begin{array}{*{35}{l}} {{a}_{12}}~=\text{ }\left( 3/2 \right){{\left( 2 \right)}^{12\text{ }\text{ }1}} \\ =\text{ }\left( 3/2 \right)\text{ }\times \text{ }{{\left( 2 \right)}^{11}} \\ =\text{ }\left( 3/2 \right)\text{ }\times \text{ }2048 \\ =\text{ }3072 \\ {{a}_{8}}~=\text{ }3072 \\ \end{array}\]

  1. In a GP., the third term is

        \[24\]

    and

        \[{{6}^{th}}\]

     term is

        \[192\]

    . Find the

        \[{{10}^{th}}\]

     term.

Solution:-

From the question it is given that,

    \[\begin{array}{*{35}{l}} {{a}_{3}}~=\text{ }24 \\ {{a}_{6}}~=\text{ }192 \\ \end{array}\]

Then, by the formula

    \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}} {{a}_{6}}~=\text{ }a{{r}^{6\text{ }\text{ }1}} \\ 192\text{ }=\text{ }a{{r}^{6\text{ }\text{ }1}} \\ \end{array}\]

    \[192\text{ }=\text{ }a{{r}^{5}}\]

 … [equation (i)]

Now,

    \[{{a}_{3}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]

    \[24\text{ }=\text{ }a{{r}^{3\text{ }\text{ }1}}\]

    \[24\text{ }=\text{ }a{{r}^{2}}~\]

… [equation (ii)]

By dividing equation (i) by equation (ii)

    \[\begin{array}{*{35}{l}} a{{r}^{5}}/a{{r}^{2}}~=\text{ }192/24 \\ {{r}^{5\text{ }\text{ }2}}~=\text{ }8 \\ {{r}^{3}}~=\text{ }8 \\ {{r}^{3}}~=\text{ }{{2}^{3}} \\ r\text{ }=\text{ }2 \\ \end{array}\]

Now, substitute the value r in equation (i),

    \[\begin{array}{*{35}{l}} 192\text{ }=\text{ }a{{r}^{5}} \\ 192\text{ }=\text{ }a\text{ }{{\left( 2 \right)}^{5}} \\ a\text{ }=\text{ }192/32 \\ a\text{ }=\text{ }6 \\ \end{array}\]

So,

    \[{{a}_{10}}~=\text{ }a{{r}^{10\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}}</strong> <strong> =\text{ }a{{r}^{9}} \\</strong> <strong> =\text{ }6{{\left( 2 \right)}^{9}} \\</strong> <strong> =\text{ }6\text{ }\left( 512 \right) \\</strong> <strong> =\text{ }3072 \\</strong> <strong>\end{array}\]

i

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