$(a)\;1280\qquad(b)\;640\qquad(c)\;2560\qquad(d)\;480$

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- In a GP, $ T_{n} = ar^{n - 1}$

In a GP, $ T_{n} = ar^{n - 1}$

Given, $ T_{4} = ar^{3} = 5 , T_{8} = ar^{7} = 80$

Dividing, $\Large \frac {T_{8}}{T_{4}} = \Large \frac {ar^{7}} {ar^{3}} = \Large \frac {80} {5} $

$\implies r^{4} = 16 \implies r =2 $

Given, $ T_{4} = ar^{3} = 5 $,

substituting $ r=2 $, we get $ a = \Large\frac {5} {8} $

Now, $ T_{12} = ar^{11} = \large\frac {5} {8} X 2^{11} = 1280 $

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