
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
August 30th, 2017, 02:18 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, 02:42 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,628 Thanks: 1308  $r = \dfrac{\sqrt{2}}{2}=2^{1/2}$ $t_n = t_1 \cdot r^{n1}$ 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, 03:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1386 
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, 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 

Tags 
geometric, question, sequence 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Arithmetic Sequence & Geometric Sequence Question  Tomedb  Algebra  1  November 3rd, 2016 03:17 PM 
Geometric sequence  med1student  Algebra  3  March 5th, 2016 08:48 AM 
geometric sequence  Vince604  Algebra  6  November 15th, 2012 08:39 PM 
Geometric sequence question  CalebB  Algebra  3  February 29th, 2012 02:12 PM 
geometric sequence  Tafka  Calculus  2  May 3rd, 2010 08:48 AM 