# Find the two values of the common ratio- Geometric sequence

• chwala
In summary: I ought to have,##\dfrac {a}{243}=\dfrac{a(1-r^4)}{240}##...##r^2=\dfrac{1}{9}####r_1=\dfrac{1}{3}## and ##r_2=-\dfrac{1}{3}####r_1=\dfrac{1}{3}## and ##r_2=-\dfrac{1}{3}##
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
sequences

*kindly note that i do not have the solutions ...I was looking at this, not quite sure on what they mean by exact fractions, anyway my approach is as follows;
##\dfrac {a}{243}=\dfrac{a(1-r^3)}{240}##

##\dfrac {1}{243}=\dfrac{1-r^3}{240}##

##\dfrac {240}{243}=1-r^3##

##r^3=1-\dfrac{240}{243}##

##r^3=\dfrac{3}{243}##

##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##

My alternative approach,
##(243-243r)(1-r^3)=240-240r##
##243r^4-243r^3-3r+3=0## using ti- nspire i realized same value as first approach...

I am not sure of my answer, I will need to check the r value and see whether i will realize the sum of ##240##. On quick checking i seem to have the correct terms i.e ##[184.4..., 43.6333..., 9.85468..., 2.2813...]## whose ##S_{4}≅240.##
I think i got it...let me post my working...

From
##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##,

We shall have;

##r_{1}=\dfrac{1}{3}## and ##r_{2}=\dfrac{1}{\sqrt [3]3}##

On checking with, ##r=\dfrac{1}{3}## , we have the terms of the sequence as ##[162,54,18,6 ]## which add up to ##240##. Cheers guys.

Last edited:
$$a_0+a_1+a_2+a_3+... =a_0\frac{1}{1-r}=\frac{1}{243}$$
$$a_0+a_1+a_2+a_3=a_0\frac{1-r^4}{1-r}=\frac{1}{240}$$
give ##r^2=1/9##.

Last edited:
chwala
anuttarasammyak said:
$$a_0+a_1+a_2+a_3+... =a_0\frac{1}{1-r}=\frac{1}{243}$$
$$a_0+a_1+a_2+a_3=a_0\frac{1-r^4}{1-r}=\frac{1}{240}$$
give ##r^2=1/9##.
Does that mean my approach is wrong? Is my ##r_2## correct? Of course, we only have one ##r## value that will satisfy our problem. Thanks @anuttarasammyak

Last edited:
chwala said:
not quite sure on what they mean by exact fractions
Exact fraction: 1/3
Not exact fraction: .333
IOW, not numbers like 9.85468...
chwala said:
##\dfrac {a}{243}=\dfrac{a(1-r^3)}{240}##
I believe you're off here. The numerator on the right side should be ##a(1 - r^4)##. If you fix that error, you get what @anuttarasammyak got, ##r^2 = \frac 1 9##, which has two solutions.
chwala said:
Does that mean my approach is wrong?
Yes, see above.

chwala
Mark44 said:
Exact fraction: 1/3
Not exact fraction: .333
IOW, not numbers like 9.85468...

I believe you're off here. The numerator on the right side should be ##a(1 - r^4)##. If you fix that error, you get what @anuttarasammyak got, ##r^2 = \frac 1 9##, which has two solutions.

Yes, see above.
silly me...was fixated on using nth term formula ...##r^{n-1}## instead of sum... ##r^n##...i did this in a rush...cheers Mark.

chwala said:
Homework Statement:: See attached.
Relevant Equations:: sequences

View attachment 301873

*kindly note that i do not have the solutions ...I was looking at this, not quite sure on what they mean by exact fractions, anyway my approach is as follows;
##\dfrac {a}{243}=\dfrac{a(1-r^3)}{240}##

##\dfrac {1}{243}=\dfrac{1-r^3}{240}##

##\dfrac {240}{243}=1-r^3##

##r^3=1-\dfrac{240}{243}##

##r^3=\dfrac{3}{243}##

##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##

My alternative approach,
##(243-243r)(1-r^3)=240-240r##
##243r^4-243r^3-3r+3=0## using ti- nspire i realized same value as first approach...

I am not sure of my answer, I will need to check the r value and see whether i will realize the sum of ##240##. On quick checking i seem to have the correct terms i.e ##[184.4..., 43.6333..., 9.85468..., 2.2813...]## whose ##S_{4}≅240.##
I think i got it...let me post my working...

From
##r=\dfrac{1}{\sqrt [3]81}=\dfrac{1}{3\sqrt [3]3}##,

We shall have;

##r_{1}=\dfrac{1}{3}## and ##r_{2}=\dfrac{1}{\sqrt [3]3}##

On checking with, ##r=\dfrac{1}{3}## , we have the terms of the sequence as ##[162,54,18,6 ]## which add up to ##240##. Cheers guys.

...I ought to have,

##\dfrac {a}{243}=\dfrac{a(1-r^4)}{240}##

...##r^2=\dfrac{1}{9}##

##r_1=\dfrac{1}{3}## and ##r_2=-\dfrac{1}{3}##

chwala said:
##r_1=\dfrac{1}{3}## and ##r_2=-\dfrac{1}{3}##
Be sure to check these, if you haven't done so already.

chwala

## 1. What is a geometric sequence?

A geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant value called the common ratio. For example, in the sequence 2, 6, 18, 54, the common ratio is 3.

## 2. How do I find the common ratio in a geometric sequence?

To find the common ratio in a geometric sequence, divide any term by the previous term. The resulting quotient will be the common ratio. For example, in the sequence 2, 6, 18, 54, the common ratio is 6/2 = 3.

## 3. What if the common ratio is a fraction or decimal?

The common ratio can be any real number, including fractions and decimals. Just follow the same method of dividing any term by the previous term to find the common ratio.

## 4. Can there be more than one common ratio in a geometric sequence?

No, there can only be one common ratio in a geometric sequence. If there are multiple common ratios, then it is not considered a geometric sequence.

## 5. How do I use the common ratio to find the next term in a geometric sequence?

To find the next term in a geometric sequence, multiply the previous term by the common ratio. For example, if the common ratio is 3 and the previous term is 18, the next term would be 18 x 3 = 54.

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