## anonymous one year ago The population of a local species of dragonfly can be found using an infinite geometric series where a1 = 42 and the common ratio is three fourths. Write the sum in sigma notation and calculate the sum that will be the upper limit of this population.

1. anonymous

@jim_thompson5910

2. jim_thompson5910

what do you have so far?

3. anonymous

i have|dw:1442973187600:dw|

4. jim_thompson5910

you're very close

5. anonymous

oh and the 42(3/4)^i-1?

6. jim_thompson5910

it should be $\Large \sum_{i=1}^{\infty} 42\left(\frac{3}{4}\right)^{i-1}$

7. anonymous

Yes! That's what I meant. Thank-you :)

8. jim_thompson5910

now calculate the sum that will be the upper limit of this population.

9. anonymous

Hmmm...it wouldn't be diveregent right?

10. anonymous

divergent*

11. jim_thompson5910

you see how a = 42 and r = 3/4 right?

12. anonymous

yes

13. anonymous

ok! So we are plugging it into the geometric series formula!

14. jim_thompson5910

yeah since |r| < 1 is true, we can use S = a/(1-r)

15. anonymous

ok. So I got 168?

16. jim_thompson5910

me too

17. anonymous

Can you help me on another?

18. jim_thompson5910

sure

19. anonymous

What is the sum of the arithmetic sequence 8, 14, 20 …, if there are 24 terms?

20. jim_thompson5910

first term = ?? common difference = ??

21. anonymous

A1=8 C.D.=-6

22. jim_thompson5910

C.D is +6 actually

23. jim_thompson5910

we add 6 to each term to get the next one

24. anonymous

oh! Sorry :)

25. jim_thompson5910

now use the formula $\Large S_n = \frac{n(2a+d(n-1))}{2}$ n = number of terms added up a = first term d = common difference

26. anonymous

So far, I have s24=24(16+6n-1/2)

27. jim_thompson5910

n = 24, so replace the n in n-1 with 24

28. anonymous

oh! Thanks

29. jim_thompson5910

$\Large S_n = \frac{n(2a+d(n-1))}{2}$ $\Large S_{24} = \frac{24(2*8+6(24-1))}{2}$ $\Large S_{24} = ???$

30. anonymous

Ok! Hold on lol

31. anonymous

1848 :)

32. jim_thompson5910

good