From the question it is given that,
Last term = 12288
First term a = 3,
Then,
![\[\begin{array}{*{35}{l}} r\text{ }=\text{ }\left( -6 \right)\text{ }\div \text{ }\left( 3 \right) \\ r\text{ }=\text{ }\left( -6 \right)\text{ }\times \text{ }\left( 1/3 \right) \\ r\text{ }=\text{ }\left( -6\text{ }\times \text{ }1 \right)/\left( 1\text{ }\times \text{ }3 \right) \\ r\text{ }=\text{ }\left( -2\text{ }\times \text{ }1 \right)/\left( 1\text{ }\times \text{ }1 \right) \\ r\text{ }=\text{ }-2 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6f5da4a817e2ef7c4b233789665528d3_l3.png "Rendered by QuickLaTeX.com")
Then, 6th term from the end,
![\[\begin{array}{*{35}{l}}</strong> <strong> {{a}_{6}}~=\text{ }l\text{ }\times \text{ }{{\left( 1/r \right)}^{n\text{ }\text{ }1}} \\</strong> <strong> =\text{ }12288\text{ }\times \text{ }{{\left( 1/-2 \right)}^{6\text{ }\text{ }1}} \\</strong> <strong> =\text{ }12288\text{ }\times \text{ }(1/-{{2}^{5}}) \\</strong> <strong> =\text{ }12288/-32 \\</strong> <strong> =\text{ }\text{ }384 \\</strong> <strong>\end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2456125cbd911548ef5b283188d99648_l3.png "Rendered by QuickLaTeX.com")
(vii) Find the 6th term from the end of the list of numbers
![\[\mathbf{3},\text{ }\text{ }\mathbf{6},\text{ }\mathbf{12},\text{ }\text{ }\mathbf{24},\text{ }\ldots ,\text{ }\mathbf{12288}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c994267795992e20cd066d203934c28d_l3.png "Rendered by QuickLaTeX.com")
From the question it is given that,
Last term = 12288
First term a = 3,
Then,
Then, 6th term from the end,
