v__r Posted November 3, 2012 Report Share Posted November 3, 2012 (edited) I'm stuck in this question i really really need to know how to solve it!The sum, Sn, of the first n terms of a geometric sequence, whose nth term is un, is given bySn = 7^(n -1) -a^(n)/7^n, where a > 0.(a) Find an expression for un.Un= 7^(n -1) -a^(n)/7^n - 7^(n-1) -a^(n-1) /7^n(b) Find the first term and common ratio of the sequence.????© Consider the sum to infinity of the sequence.(i) Determine the values of a such that the sum to infinity exists.????(ii) Find the sum to infinity when it exists.????Thank you in advance! Edited November 3, 2012 by v__r Reply Link to post Share on other sites More sharing options...
macrofire Posted November 3, 2012 Report Share Posted November 3, 2012 plug in 1 in Sn to get the 1st term. plug in 2 and do some math to get the common ratio of the sequence. Now use your new found knowledge to recreate the sum of this geometric sequence by using a formula. Evaluate the values of "a" that makes this series converge. Now find its infinite sum. Reply Link to post Share on other sites More sharing options...
flinquinnster Posted November 3, 2012 Report Share Posted November 3, 2012 I think this is an IB past paper question, so if you're really stuck and can't follow what macrofire said, then you can look it up online But you should be fine though, just remember to substitute. Reply Link to post Share on other sites More sharing options...
Meat Posted November 6, 2012 Report Share Posted November 6, 2012 Sometimes the first term is written as a, but this is not the case here. That cost me quite a bit of marks when we were assigned this problem, so try not to do the same mistake as I did.For finding the first term just substitute n with one for either Sn or Un, you get the same answer.Now to find the common ratio you can just divide Un by Un-1, (n-1)th term and you should get an answer, since the common ratio can be gotten simply by dividing any term by the previous term. You can also just divide the 2nd term by the 1st term like one of the above posts said.To find values of a such that the sum to infinity exist you should remember that for a sum to infinity to exist the common ratio should be between -1 and 1, not inclusive. So just take the common ratio and put it in this format: -1<r<1, where r is the ratio. Use the ratio you got before and substitute it into this inequality, and solve for a. Instead of writing -x<a<x, just write 0<a<x because the question states that a is larger than zero, so it cannot be negative.To find the sum of infinity just use the formula/equation: (U1)/(1-r) and get the answer. The sum to infinity should be 1 if you did everything correct. Reply Link to post Share on other sites More sharing options...
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