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Given that the fourth and eighth term of a GP are 5 and 80, find the 12th term.

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

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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 $

 

answered Nov 19, 2013 by harini.tutor
 

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