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August 30th, 2017, 02:18 AM   #1
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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
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 August 30th, 2017, 02:42 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,885 Thanks: 1504 $r = \dfrac{\sqrt{2}}{2}=2^{-1/2}$ $t_n = t_1 \cdot r^{n-1}$ note $t_1=8=2^3$ and $r^{n-1} = (2^{-1/2})^{n-1} = 2^{1/2-n/2}$ $t_n= 2^3 \cdot 2^{1/2-n/2} = 2^{3 + 1/2 - n/2} = 2^{7/2 -n/2} = 2^{(7-n)/2}$ Thanks from maths98765
 August 30th, 2017, 03:44 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,474 Thanks: 2039 If a geometric sequence has common ratio $r$ and first term $a$, its $n$th term is $ar^{\large n-1}\!$. For the above problem, that's 8(√2/2)$^{\large n-1}\!$ = 2$^{\large3 + (1/2)(n - 1) - (n - 1)}\!$ = 2$^{\large\frac72 - \frac12n}\!$. Thanks from maths98765
 August 31st, 2017, 06:26 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 Thanks from maths98765 and verygoodatmaths

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