How Many Terms Are in the Geometric Sequence?

[chikis]

Homework Statement

How many terms
has the G.p whose
2nd term is 1/2
and common ratio
and the last term
1/4 and 1/128
respectiveley?

From the question, it can be deduced that the common ratio, r =1/4 and the last term = 1/128

Homework Equations

The N_t term of a G.p =
ar^n-1

The Attempt at a Solution

Given that r, the
common ratio=1/4.
T₂, the second
term is =1/2.
Using the formula
ar^n-1
where a = first
term and r is =
common ratio.
Thus:T₂=a(1/4)^2-1
=1/2
a(1/4) = 1/2
a/4 = 1/2
2a = 4
a = 4/2
a = 2
Recall that T_n of
G.p = ar^n-1
To find the N_t
term, we need to put a, r and the last term in equation
ar^n-1 = 1/128
2(1/4)^n-1 =1/128
I multiplied through
by 1/2
2 * 1/2(1/4)^n-1 =
1/128 * 1/2
(1/4)^n-1 = 1/256
I don't know how
to deal with the
power n-1 in other
to procceed
further and arive
at the answer.

 

[HallsofIvy]

Why not just "do the arithmetic"? What is 2²? What is 2³? What is 2⁴? ...

This works because this problem happens to be "set up" so that it is easy- 256 happens to be an integer power of 4. More generally to solve aⁿ= b for n, you would have to use logarithms. Taking the logarithm of both sides, log(aⁿ)= n log(a)= log(b) so n= log(b)/log(a).

You can use logarithm to whatever base is convenient.

 

[chikis]

Why not just "do the arithmetic"? What is 2²? What is 2³? What is 2⁴? ...

This works because this problem happens to be "set up" so that it is easy- 256 happens to be an integer power of 4. More generally to solve aⁿ= b for n, you would have to use logarithms. Taking the logarithm of both sides, log(aⁿ)= n log(a)= log(b) so n= log(b)/log(a).

You can use logarithm to whatever base is convenient.

Logarithm! Hm! I have never seen where logarithm is aplied in problem relating to G.p such at this; atleast not in my level. Anyway, am not afraid to learn new things; maybe you should start while I watch and learn. No abuse intended, am just expressing my mind.

 

[Chestermiller]

Homework Statement

How many terms
has the G.p whose
2nd term is 1/2
and common ratio
and the last term
1/4 and 1/128
respectiveley?

2, 1/2, 1/8, 1/32, 1/128 (5 terms)

 

[chikis]

2, 1/2, 1/8, 1/32, 1/128 (5 terms)

Thank you but I must add this:
the answer is not the problem but how or what procedure did you employ to arrive at that answer; that is my main problem.

 

[Chestermiller]

Thank you but I must add this:
the answer is not the problem but how or what procedure did you employ to arrive at that answer; that is my main problem.

If the second term is 1/2, and the common ratio is 1/4, then the first term must have been (1/2)/(1/4)=2. Once you know this, you just write out the terms in order, and count the number of terms required to reach 1/128.

Chet

 

[chikis]

If the second term is 1/2, and the common ratio is 1/4, then the first term must have been (1/2)/(1/4)=2. Once you know this, you just write out the terms in order, and count the number of terms required to reach 1/128.

Chet

(1/2)/(1/4)=2

As I stated earliar, the answer is not the problem but what is the problem is how to do and get that answer. If you watch my working from the begining, you will see that the way I got the value for my first term, a, is quite different from the way you got yours. From your working you have only gotting the value for the first term. If this your working is reliable, can you use it to get the value for the third term, fourth term and finally the fifth term.

 

[Joffan]

As I stated earliar, the answer is not the problem but what is the problem is how to do and get that answer. If you watch my working from the begining, you will see that the way I got the value for my first term, a, is quite different from the way you got yours. From your working you have only gotting the value for the first term. If this your working is reliable, can you use it to get the value for the third term, fourth term and finally the fifth term.

Well, no, that would just be enumerating the series. Using the formula, you can directly infer the starting term and then the position of the given term. Working through the series is all very well in this case, and a great check on your working, but the general idea of getting a particular term in the series without doing that is more useful.

Which is what you said anyway, Chikis, and asked for help on diong that with 2.(1/4)^n-1 = 1/128

So specifically on that:
$$\begin{align} 2(\frac{1}{4})^{n-1} &= \frac{2}{4^{n-1}}\\ &= \frac{2}{2^{2n-2}}\\ &= \frac{1}{2^{2n-3}} \end{align} \\ 128 = 2^7$$
and solve for n.

 

[chikis]

Well, no, that would just be enumerating the series. Using the formula, you can directly infer the starting term and then the position of the given term. Working through the series is all very well in this case, and a great check on your working, but the general idea of getting a particular term in the series without doing that is more useful.

Which is what you said anyway, Chikis, and asked for help on diong that with 2.(1/4)^n-1 = 1/128

So specifically on that:
$$\begin{align} 2(\frac{1}{4})^{n-1} &= \frac{2}{4^{n-1}}\\ &= \frac{2}{2^{2n-2}}\\ &= \frac{1}{2^{2n-3}} \end{align} \\ 128 = 2^7$$
and solve for n.

Well, all these are not necessary, I have already gotten the solution to the problem.
Here is the solution:
am starting from where I stopped

(1/4)^n-1 = 1/256
(4^-1)^n-1 = (256^-1)
(4)^-n+1 = (2⁸)^-1
(2²)^-n+1 = (2)^-8
2^-2n+2 = 2^-8
the two equation are now equal, so we now equate the both indices of both equation and solve for n,
-2n+2 = -8
-2n = -8-2
-2n = -10
n = -10/-2
n = 5
and that is the number of terms in the G.p.
If we want to still go futher and write out the value of each terms, then we get it by mutiplying the the ratio with each term to get the next term after it:
we arleady know from the given question and and our calculation that the common ratio, r, the first term, a, the second term, T₂ and the last term, T₃ are 1/4, 2, 1/2 and 1/128; what is remaining for us now is to get the value for the third term, T₃ and the the fourth term, T₄ to get it we do the following,
T₃ = 1/2*1/4 = 1/8
T₄ = 1/8*1/4 = 1/32
:. The complete G.p is as follows
2, 1/2, 1/8, 1/32, 1/128.
Thus the problem is solved:-D.

 

[Chestermiller]

As I stated earliar, the answer is not the problem but what is the problem is how to do and get that answer. If you watch my working from the begining, you will see that the way I got the value for my first term, a, is quite different from the way you got yours. From your working you have only gotting the value for the first term. If this your working is reliable, can you use it to get the value for the third term, fourth term and finally the fifth term.

According to the definition of a geometric progression, each term in the progression is equal to the previous term times the common ratio. That's all the information I used.
Silly me. I thought you were just looking for a solution to the problem. I didn't realize it had to be done using exactly what you learned in Algebra class. Here is another lesson in algebra:

(1/4)**(n-1)=(1/256)

4**(n-1)=256=4**(4)

n-1=4

n=5

Chet

 

[chikis]

According to the definition of a geometric progression, each term in the progression is equal to the previous term times the common ratio. That's all the information I used.
Silly me. I thought you were just looking for a solution to the problem. I didn't realize it had to be done using exactly what you learned in Algebra class. Here is another lesson in algebra:

(1/4)**(n-1)=(1/256)

4**(n-1)=256=4**(4)

n-1=4

n=5

Chet

I don't understand your own working from this line:
(1/4)**(n-1)=(1/256)

4**(n-1)=256=4**(4)

Can you explain please!

 

[Chestermiller]

($\frac{1}{4}$)^(n-1)=$\frac{1}{256}$

4^(n-1)=256=4⁴

n-1=4

n=5

 

[chikis]

($\frac{1}{4}$)^(n-1)=$\frac{1}{256}$

4^(n-1)=256=4⁴

n-1=4

n=5

I don't undestand all these written in bold form:
($\frac{1}{4}$)^(n-1)=$\frac{1}{256}$
Write it in such a way that I may understand it. Don't you see my own? Is there anything you don't understand in my own working? Just put it in a simple form just like mine.

 

[Chestermiller]

Yes. To me on my computer, the equations look pretty much the same as the ones in your replies. I used the PF equation editor... I think they call it latex. I don't know what else I can do.

Chet

 

[chikis]

Yes. To me on my computer, the equations look pretty much the same as the ones in your replies. I used the PF equation editor... I think they call it latex. I don't know what else I can do.

Chet

Then how do I get the exact equation of your working?

 

[Chestermiller]

I'm going to try to write it out in words. I hope this works for you.

(1/4) to the power (n-1) is equal to (1/256).

(1/256) is equal to (1/4) to the power 4.

So, (1/4) to the power (n-1) is equal to (1/4) to the power 4.

Therefore, (n-1) = 4

n=5

 

[chikis]

I'm going to try to write it out in words. I hope this works for you.

(1/4) to the power (n-1) is equal to (1/256).

(1/256) is equal to (1/4) to the power 4.

So, (1/4) to the power (n-1) is equal to (1/4) to the power 4.

Therefore, (n-1) = 4

n=5

That's a good one. Here is what you mean in figures.
(1/4)^n-1 = (1/256)
(1/4)^n-1 = (1/4)⁴
n-1 = 4
n = 4+1
n = 5.
That's is a good one. Atleast one's brain will be sharpened. Thank you for that idea.