joeyay Posted April 5, 2013 Report Share Posted April 5, 2013 A geometric series has a first term 400, ten terms and a sum of 1295.67. What is the common ratio?I am not sure how to solve for r^10. Can anyone help me? Reply Link to post Share on other sites More sharing options...
SurajM Posted April 5, 2013 Report Share Posted April 5, 2013 Sum = term 1/ (1-common ratio) 1295.67 = 400/ (1-r) just use this formula and solve for r Hope it helps! 1 Reply Link to post Share on other sites More sharing options...
Siger Posted April 5, 2013 Report Share Posted April 5, 2013 S10 = 1295.67 Un = ar^n-1 Sn = a (1-r^n/ 1-r) Therefore, 1295.67 = 400 * ((1 - r^10) / (1 - r)) Dividing both sides by 400, 3.24 = (1-r^10)/ (1-r) Cross multiplying, 3.24 - 3.24r = 1- r^10 therefore, r^10 - 3.24 r + 2.24 = 0 I was stuck here but when graphing this equation on a graphics calculator, and u look for the x-intercepts, r=0.7 r=1 but r cannot b 1 so, r= 0.7 Checking your answer, 400 * ((1- 0.7^10)/ (1-0.7)) = 1295.66667 Dont know if i helped enough... :-) 2 Reply Link to post Share on other sites More sharing options...
SurajM Posted April 5, 2013 Report Share Posted April 5, 2013 I'm sorry, the Siger's explanation would be right.I gave you the answer if this was an infinite geometric sequence, but since it has a set number of terms you need to use the formula he used.Sorry! Reply Link to post Share on other sites More sharing options...
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