The first term of a geometric progression is 3 and then common ratio is r, where |r|<1.

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cired2002

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The first term of a geometric progression is 3 and then common ratio is r, where |r|<1. The sum of the first three terms of the progression is 8/9 of the sum of the first six terms. Find the sum to infinity.
 
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What have you tried? Where are you stuck? Exactly what do you need help with? It is hard to help you if you show us no work. In fact that is why the forum rules ask that you show your work.

To know the series you need to find r. You know the 1st three terms are a, ar, ar2 and they tell you a. So find r!
 
I would begin by computing the \(n\)th partial sum \(S_n\):

[MATH]S_n=a+ar+ar^2+\cdots+ar^{n-1}+ar^n[/MATH]
Now, suppose we multiply by \(r\)

[MATH]rS_n=ar+ar^2+ar^3+\cdots+ar^{n}+ar^{n+1}[/MATH]
Subtracting the former from the latter, we obtain:

[MATH](r-1)S_n=a\left(r^{n+1}-1\right)[/MATH]
Hence:

[MATH]S_n=\frac{a\left(r^{n+1}-1\right)}{r-1}[/MATH]
You've likely seen this formula before. As Jomo mentioned, you are given \(a\), and you are told:

[MATH]S_2=\frac{8}{9}S_5[/MATH]
From this, you may compute \(r\) without even needing to know \(a\), since it will divide out. What do you find?
 
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cired2002 said:
The first term of a geometric progression is 3 and then common ratio is r, where |r|<1. The sum of the first three terms of the progression is 8/9 of the sum of the first six terms. Find the sum to infinity.
Click to expand...
Have a look at this. Now we know that if \(\displaystyle |r|<1\) then \(\displaystyle \sum\limits_{k = J}^\infty {a{r^k}} = \frac{{a{r^J}}}{{1 - r}}\)
 
Jomo said:
What have you tried? Where are you stuck? Exactly what do you need help with? It is hard to help you if you show us no work. In fact that is why the forum rules ask that you show your work.

To know the series you need to find r. You know the 1st three terms are a, ar, ar2 and they tell you a. So find r!
Click to expand...
I used the formula l. I put S3 = 8/9 S6 and I sub in a=3. But I can't get the answer
 
MarkFL said:
I would begin by computing the \(n\)th partial sum \(S_n\):

[MATH]S_n=a+ar+ar^2+\cdots+ar^{n-1}+ar^n[/MATH]
Now, suppose we multiply by \(r\)

[MATH]rS_n=ar+ar^2+ar^3+\cdots+ar^{n}+ar^{n+1}[/MATH]
Subtracting the former from the latter, we obtain:

[MATH](r-1)S_n=a\left(r^{n+1}-1\right)[/MATH]
Hence:

[MATH]S_n=\frac{a\left(r^{n+1}-1\right)}{r-1}[/MATH]
You've likely seen this formula before. As Jomo mentioned, you are given \(a\), and you are told:

[MATH]S_3=\frac{8}{9}S_6[/MATH]
From this, you may compute \(r\) without even needing to know \(a\), since it will divide out. What do you find?
Click to expand...
I got r = 29/16 and r = 1 and both will not lead me to my answer
 
Seriously if you don't show us your work we can't determine where you are making your mistakes. If you show us your work you will get expert help. Please play by the rules of the forum (did you read them) and show us your work.
 
cired2002 said:
I got r = 29/16 and r = 1 and both will not lead me to my answer
Click to expand...

edit: I should have written earlier that:

[MATH]S_2=\frac{8}{9}S_5[/MATH]
because we begin counting with the zeroth term. I apologize for the confusion.

To find \(r\), I would write:

[MATH]\frac{a\left(r^{2+1}-1\right)}{r-1}=\frac{8}{9}\cdot\frac{a\left(r^{5+1}-1\right)}{r-1}[/MATH]
Multiply through by [MATH]\frac{9(r-1)}{a}[/MATH]:

[MATH]9\left(r^{3}-1\right)=8\left(r^{6}-1\right)[/MATH]
[MATH]8r^6-9r^3+1=0[/MATH]
If we recognize this is q quadratic in \(r^3\), we may factor as follows:

[MATH]\left(8r^3-1\right)\left(r^3-1\right)=0[/MATH]
Given that \(|r|<1\), the only real root will come from:

[MATH]8r^3-1=0\implies r=\frac{1}{2}[/MATH]
Now, we should see that:

[MATH]S_{\infty}=\frac{a}{1-r}[/MATH]
Since:

[MATH]\lim_{n\to\infty}S_n=\frac{a}{1-r}[/MATH]
And so what do you get for the infinite sum of the given series?
 
8
MarkFL said:
edit: I should have written earlier that:

[MATH]S_2=\frac{8}{9}S_5[/MATH]
because we begin counting with the zeroth term. I apologize for the confusion.

To find \(r\), I would write:

[MATH]\frac{a\left(r^{2+1}-1\right)}{r-1}=\frac{8}{9}\cdot\frac{a\left(r^{5+1}-1\right)}{r-1}[/MATH]
Multiply through by [MATH]\frac{9(r-1)}{a}[/MATH]:

[MATH]9\left(r^{3}-1\right)=8\left(r^{6}-1\right)[/MATH]
[MATH]8r^6-9r^3+1=0[/MATH]
If we recognize this is q quadratic in \(r^3\), we may factor as follows:

[MATH]\left(8r^3-1\right)\left(r^3-1\right)=0[/MATH]
Given that \(|r|<1\), the only real root will come from:

[MATH]8r^3-1=0\implies r=\frac{1}{2}[/MATH]
Now, we should see that:

[MATH]S_{\infty}=\frac{a}{1-r}[/MATH]
Since:

[MATH]\lim_{n\to\infty}S_n=\frac{a}{1-r}[/MATH]
And so what do you get for the infinite sum of the given series?
[/QU
MarkFL said:
edit: I should have written earlier that:

[MATH]S_2=\frac{8}{9}S_5[/MATH]
because we begin counting with the zeroth term. I apologize for the confusion.

To find \(r\), I would write:

[MATH]\frac{a\left(r^{2+1}-1\right)}{r-1}=\frac{8}{9}\cdot\frac{a\left(r^{5+1}-1\right)}{r-1}[/MATH]
Multiply through by [MATH]\frac{9(r-1)}{a}[/MATH]:

[MATH]9\left(r^{3}-1\right)=8\left(r^{6}-1\right)[/MATH]
[MATH]8r^6-9r^3+1=0[/MATH]
If we recognize this is q quadratic in \(r^3\), we may factor as follows:

[MATH]\left(8r^3-1\right)\left(r^3-1\right)=0[/MATH]
Given that \(|r|<1\), the only real root will come from:

[MATH]8r^3-1=0\implies r=\frac{1}{2}[/MATH]
Now, we should see that:

[MATH]S_{\infty}=\frac{a}{1-r}[/MATH]
Since:

[MATH]\lim_{n\to\infty}S_n=\frac{a}{1-r}[/MATH]
And so what do you get for the infinite sum of the given series?
Click to expand...
omg I totally got it now. It turned out I subs in r as a in the denominator in the equation. Thank you so much for your help!!appreciate it!
Click to expand...
 
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