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November 26, 2014

November 26, 2014

Posted by **Anonymous** on Sunday, October 23, 2011 at 1:06pm.

- Geometric progression -
**Charles**, Friday, February 17, 2012 at 10:26amSolution.Tn=ar'n-1 where n=3

- Geometric progression -
**Babalola Joshua**, Thursday, October 11, 2012 at 7:55amar = 12 + a (i)

ar^2 = x (ii)

subtract equ (i) from equ (ii)

a = 6

substituting a = 6 into equ (i)

ar = 12 + a

6r = 12 + 6=18

r = 3.

ar'2 = 6 * 3'2

ar'2 = 54

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