|
04-20-2011, 06:00 PM | #1 |
FFR Veteran
Join Date: Nov 2010
Posts: 2,040
|
Math [Sequences and Series] Question! Grade 11. Help please !!
I have a biology summative due tomorrow, an English test and a crapload of physics homework that'll take me a long time to finish. I missed 2 classes of Math and I don't have time to learn this so someone just please solve these 2 questions and i'll be very happy!
Both number 4 and 5 (ignore the fact that it's crossed out)
__________________
I have a dig bick. |
04-20-2011, 06:14 PM | #2 |
Hunger Games Hunty
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
4. t4 = 45
t6 = 180 A geometric series has an r that describes the ratio between each successive term. Dividing 180 by 45 gives us a ratio of 4. We must split this by 2 again to account for the t5. Thus, 2 is the geometric ratio. Taking 45 and diving it by this ratio 3 times will bring you back to t1. 45/(2)^3 gives you 45/8, or the first term (a) of the series. An = A(R)^(n-1) So I get: Tn = (45/8)(2)^(n-1) If no one helps you out with 5 I'll stop by later. I'm helping my sister with math right now, hahahaha. Last edited by Jtehanonymous; 04-20-2011 at 06:17 PM.. |
04-20-2011, 06:18 PM | #3 |
FFR Veteran
Join Date: Nov 2010
Posts: 2,040
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
Thanks, I understand number 4 now
__________________
I have a dig bick. |
04-20-2011, 06:20 PM | #4 |
D6 FFR Legacy Player
Join Date: Jan 2009
Age: 33
Posts: 4,342
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
Recall that a geometric sequence is a sequence of elements found by multiplying some constant onto the initial term recursively. In mathematical notation, if a is the constant and x is the initial term, the following is a geometric series:
x , ax , a^2 x , a^3 x , ... You are given the fourth and the sixth term of the sequence, so you know that x_6 = a^2 x_4, because you need to multiply a twice to get from the fourth to the sixth term. In other words: 180 = a^2 (45) Or a = 2. Now all we need to do is find the initial term. x_0 a^4 = x_4 because you need to multiply a four times to get from the first to the fourth term. This implies that x_0 (2^4) = 45 or, x_1 = 45/16. Now we just need to create a formula to express all t_n. Starting term is t_0 = 45/16, so t_n = 45/16 (something). That something is a^n because to get to t_n, you need to multiply a onto t_0 n times. Therefore t_n = 45/16 (2)^n EDIT: Ninja'd >8( |
04-20-2011, 06:22 PM | #5 |
Hunger Games Hunty
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
Wilson, I'm still suing you for 25,000 credits for that one time.
I thought you should know~ All pro: No problem! |
04-20-2011, 06:26 PM | #6 |
FFR Veteran
Join Date: Nov 2010
Posts: 2,040
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
lol wilson, thanks for the help anyways.
__________________
I have a dig bick. |
04-20-2011, 06:28 PM | #7 | |
D6 FFR Legacy Player
Join Date: Jan 2009
Age: 33
Posts: 4,342
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
Question 2:
Notice that when t_1 = 2, then t_2 = t_1 - 3 = -2-3 = -5. Also, t_3 = t_2 - 3 = (t_1 - 3) - 3 = t_1 - (2)3 = -2 - 6 = -8 And so on: t_n = t_(n-1) - 3 = (t_(n-2) - 3) - 3 = (t_(n-2) - (2)(3)) = (t_(n-3) - 3) - 2(3) = (t_(n-3) - 3(3) = .... = t_(n- (n-1)) - 3(n-1) = t_1 -3(n-1) So: t_n = -2 - 3(n-1) Let me know if this is too confusing lol EDIT: Quote:
and All-Pro, no problem :] |
|
04-20-2011, 07:07 PM | #8 | |
FFR Veteran
Join Date: Nov 2010
Posts: 2,040
|
Re: Math [Sequences and Series] Question! Grade 11. Help please !!
Quote:
__________________
I have a dig bick. |
|
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | |
Thread Tools | |
Display Modes | |
|
|