## anonymous one year ago Hello! I don't know why I am getting the wrong answer, please help! 10 .The first three terms of a geometric sequence are 100, 90 and 81. (iv) After how many terms is the sum of the sequence greater than 99% of the sum to infinity. I get -22. the answer according to the text book is 44. In a previous question it is shown that the r=9/10

1. anonymous

oh I forgot to mention the sum to infinity of the terms of the sequence is 1000

2. DanJS

Did you figure out the general sequence yet using those terms given

3. DanJS

oh ok, had to look what that meant, sum to infinity is the limit of that sequence

4. DanJS

have you seen this before: $S _{n} = \frac{ a _{1}(1-r^n) }{ 1-r }$ sum of n terms of geometric series

5. DanJS

r = ratio of n+1 term to the nth term ... r = 90/100 = 81/90 = 0.9

6. DanJS

since the terms are always getting smaller and smaller by that ratio,, eventually the thing will settle or 'converge' to a certain value, the farther out you go in the value of n, the closer the total will be to the limiting value of the series

7. DanJS

sorry back,

8. DanJS

yeah 44 looks right

9. DanJS

since they just gave you the sum to infinity is 1000, 99% of that is 990 you want that series to total 990 or more and find out how many terms it takes, n

10. DanJS

$990=\frac{ 100(1-0.9^n )}{ 1-0.9 }$ solve

11. DanJS

you see the value for n, round up to the next whole number, that is the number of terms to get to 990, or , 99% of 1000

12. DanJS

i got 43.something, so 44 terms to get there

13. anonymous

Oh I think I see. Thank you very much!

14. DanJS

yes yes, if they dont give you a value for the limit or infinite sum, then you can find it by the first value in the sequence divided by (1 - r) 100/(1-0.9) = 1000

15. DanJS

if r is larger than 1, then the thing will not have an infinite sum or converging value as n goes to larger

16. DanJS

in that case it just blows up since each term is larger than the last

17. anonymous

Ok. Good to know. :) Thanks very much!