T Twickel Member Joined Dec 5, 2007 Messages 390 Gender Male HSC 2009 Nov 12, 2008 #1 Hi How do I find he common ration for Q7 a (i) Its 2/root2-1/2 sohw working please. I keep getting Root2+1/1 but it should be the inverse of that 1/root+1 WHY?
Hi How do I find he common ration for Q7 a (i) Its 2/root2-1/2 sohw working please. I keep getting Root2+1/1 but it should be the inverse of that 1/root+1 WHY?
V vds700 Member Joined Nov 9, 2007 Messages 860 Location Sydney Gender Male HSC 2008 Nov 12, 2008 #2 The common ratio is 1/(sqrt2 + 1) Now to find the limiting sum, use the formula S = a/(1-r) =2/[1-(1/(sqrt2 +1))] =2/[(sqrt2 + 1 - 1)/(sqrt2 + 1)] =2/[sqrt2/(sqrt2 + 1)] =2(sqrt2 + 1)/sqrt2 =sqrt2(sqrt2 + 1) =2 + sqrt2 Last edited: Nov 12, 2008
The common ratio is 1/(sqrt2 + 1) Now to find the limiting sum, use the formula S = a/(1-r) =2/[1-(1/(sqrt2 +1))] =2/[(sqrt2 + 1 - 1)/(sqrt2 + 1)] =2/[sqrt2/(sqrt2 + 1)] =2(sqrt2 + 1)/sqrt2 =sqrt2(sqrt2 + 1) =2 + sqrt2
S Shoom Member Joined Aug 26, 2008 Messages 694 Gender Undisclosed HSC N/A Nov 12, 2008 #3 I need help finding the common ratio, also can you help with http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2004exams/pdf_doc/maths_04.pdf question 7c thankyou.
I need help finding the common ratio, also can you help with http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2004exams/pdf_doc/maths_04.pdf question 7c thankyou.
V vds700 Member Joined Nov 9, 2007 Messages 860 Location Sydney Gender Male HSC 2008 Nov 12, 2008 #4 I found the common ratio by inspection, however u can find it by dividing one term by the previous ter, e.g [2/(sqrt2 + 1)^2]/[2/(sqrt2 + 1) =1/(sqrt2 + 1) once u simplify down a bit
I found the common ratio by inspection, however u can find it by dividing one term by the previous ter, e.g [2/(sqrt2 + 1)^2]/[2/(sqrt2 + 1) =1/(sqrt2 + 1) once u simplify down a bit