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August 30th, 2017, 03:18 AM  #1 
Newbie Joined: Aug 2017 From: New Zealand Posts: 4 Thanks: 4  Geometric Sequence Question
Someone pls explain how you arrive at this general term. If the common ratio is root2/2 and the first term is 8, I cannot derive the general term. Thanks

August 30th, 2017, 03:42 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,805 Thanks: 1449  $r = \dfrac{\sqrt{2}}{2}=2^{1/2}$ $t_n = t_1 \cdot r^{n1}$ note $t_1=8=2^3$ and $r^{n1} = (2^{1/2})^{n1} = 2^{1/2n/2}$ $t_n= 2^3 \cdot 2^{1/2n/2} = 2^{3 + 1/2  n/2} = 2^{7/2 n/2} = 2^{(7n)/2}$ 
August 30th, 2017, 04:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
If a geometric sequence has common ratio $r$ and first term $a$, its $n$th term is $ar^{\large n1}\!$. For the above problem, that's 8(√2/2)$^{\large n1}\!$ = 2$^{\large3 + (1/2)(n  1)  (n  1)}\!$ = 2$^{\large\frac72  \frac12n}\!$. 
August 31st, 2017, 07:26 PM  #4 
Newbie Joined: Jun 2017 From: Perth, Australia Posts: 8 Thanks: 7 
To prove it you need to find a common ratio. To do so take the t2 and divide by t1 and show that this is equal to the t3 divided by the t2. Hope this helps, and if you do this, you'll find that it gives a common ratio of root two on two 

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geometric, question, sequence 
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