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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
Attached Images Screen Shot 2017-08-30 at 6.10.42 PM.jpg (7.5 KB, 25 views) August 30th, 2017, 02:42 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $r = \dfrac{\sqrt{2}}{2}=2^{-1/2}$ $t_n = t_1 \cdot r^{n-1}$ 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: 20,921 Thanks: 2203 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 Tags geometric, question, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Tomedb Algebra 1 November 3rd, 2016 03:17 PM med1student Algebra 3 March 5th, 2016 08:48 AM Vince604 Algebra 6 November 15th, 2012 08:39 PM CalebB Algebra 3 February 29th, 2012 02:12 PM Tafka Calculus 2 May 3rd, 2010 08:48 AM

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