Math [Sequences and Series] Question! Grade 11. Help please !!

[xXAll-ProXx]

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)

[Jtehanonymous]

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.

[xXAll-ProXx]

Re: Math [Sequences and Series] Question! Grade 11. Help please !!

Thanks, I understand number 4 now

[iironiic]

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(

[Jtehanonymous]

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!

[xXAll-ProXx]

Re: Math [Sequences and Series] Question! Grade 11. Help please !!

lol wilson, thanks for the help anyways.

[iironiic]

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:

Wilson, I'm still suing you for 25,000 credits for that one time.

I thought you should know~

All pro: No problem!

>8(

and All-Pro, no problem :]

[xXAll-ProXx]

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:



>8(

and All-Pro, no problem :]

Thank you Wilson, and Jteh, it wasn't as complicated as i thought this section would be :P