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(vii) Find the 6th term from the end of the list of numbers

    \[\mathbf{3},\text{ }\text{ }\mathbf{6},\text{ }\mathbf{12},\text{ }\text{ }\mathbf{24},\text{ }\ldots ,\text{ }\mathbf{12288}\]

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Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
(vii) Find the 6th term from the end of the list of numbers

    \[\mathbf{3},\text{ }\text{ }\mathbf{6},\text{ }\mathbf{12},\text{ }\text{ }\mathbf{24},\text{ }\ldots ,\text{ }\mathbf{12288}\]

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From the question it is given that,

Last term = 12288

First term a = 3,

Then,

    \[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( -6 \right)\text{ }\div \text{ }\left( 3 \right) \\ r\text{ }=\text{ }\left( -6 \right)\text{ }\times \text{ }\left( 1/3 \right) \\ r\text{ }=\text{ }\left( -6\text{ }\times \text{ }1 \right)/\left( 1\text{ }\times \text{ }3 \right) \\ r\text{ }=\text{ }\left( -2\text{ }\times \text{ }1 \right)/\left( 1\text{ }\times \text{ }1 \right) \\ r\text{ }=\text{ }-2 \\ \end{array}\]

Then, 6th term from the end,

    \[\begin{array}{*{35}{l}}</strong> <strong> {{a}_{6}}~=\text{ }l\text{ }\times \text{ }{{\left( 1/r \right)}^{n\text{ }\text{ }1}} \\</strong> <strong> =\text{ }12288\text{ }\times \text{ }{{\left( 1/-2 \right)}^{6\text{ }\text{ }1}} \\</strong> <strong> =\text{ }12288\text{ }\times \text{ }(1/-{{2}^{5}}) \\</strong> <strong> =\text{ }12288/-32 \\</strong> <strong> =\text{ }\text{ }384 \\</strong> <strong>\end{array}\]

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