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November 17th, 2013, 09:48 PM  #1 
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0  Several questions on Arithmetic and Geometric progression
i hope i didn't make mistakes in presenting my answer for that wrongs steps are given no mark even if the answer is right.... since my answer sheet show no workings at all , i can only ask for help here. 1) Simplify altho i write it as , but i don't know why.... but > can someone show me how to do simplify this? i think it is related to something like long division....but can anyone show me the steps and workings for these kind of simplification? after all the question ask to find common ratio by just giving , c is a given constant. (this is not related to ap and gp , but i would just like to know how to operate these types of equation for that it will come back in polynomial... 2) Finding Sn is minimum.. for an Arithmetic Progression Given , find when it is minimum... for AP , the turning point only has 1 so finding either or should do fine right? . . . or 3) Given a geometric progression's first 3 terms , find the common ratio and the n value when the difference between S_{infinity} and S_{n} is less than \frac{1}{10^5} common ratio , since r<1 , a=2 , S_{infinity}=\frac{a}{1r}=\frac{8}{5} }" /> }" /> square both side , }]^2" /> }]^2" /> }]^2" /> }]^2}{log 0.25}" /> because log cannot accept negative numbers....i squared both sides and ended up changing the original r value in order to get the answer.....or the actual method is different? 
November 18th, 2013, 07:26 AM  #2  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Several questions on Arithmetic and Geometric progressio Quote:
Quote:
Quote:
You know that a square is never negative so the quantity in the parentheses is never lower than . Quote:
Quote: Rather, you know that the first three numbers of a geometric sequence are a, ar, and . Since you are told that the first three numbers are 2, 1/2, and 1/8, you know that a= 2 and ar= 1/2. Since a= 2, 2r= 1/2. What is r? Use the third term, 1/8, as a check. While I don't understand your logic, you have got the correct values. Quote:
There is no need to square both sides. For n odd, the left side will be negative so necessarily less than the right side which is positive. If n is even, [latex](1/4)^n= (1/4)^n[latex]. So you must have }" />. Taking the (common) logarithm of both sides, " />. Then so [latex]n= \dfrac{5.2049}{.60206}= 8.646[/tex] so that the smallest n that works (rounding up to an integer) is 9. It is easy to check. with n= 9, while [tex]\frac{2}{1 (1/4)}= \frac{2}{5/4}= 8/5= 1.6. The difference is 0.000006... which is less than .  
November 18th, 2013, 01:34 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,823 Thanks: 723  Re: Several questions on Arithmetic and Geometric progressio
1) 1x² = (1x)(1+x). x = r^5.

November 18th, 2013, 08:19 PM  #4 
Senior Member Joined: Apr 2010 Posts: 128 Thanks: 0  Re: Several questions on Arithmetic and Geometric progressio oh yea , he did taught that..... __________________________________________________ _ i made a typo.....its this is to made sure it is a geometric progression altho the question alrd says it is a GP. the end value would get 0.25 = 0.25 r = r r = 0.25 __________________________________________________ __ So , i just need to write this as opening and start finding the n value.. , when n = even when n is even ,}" /> . . . n > blablabla.. therefore minimum value of n for}" /> is n=9 (because n is even when n > blablabla) 

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