Find the two values of the common ratio- Geometric sequence

[chwala]

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.

 

[anuttarasammyak]

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

 

[chwala]

$$a_0+a_1+a_2+a_3+... =a_0\frac{1}{1-r}=\frac{1}{243}$$
$$a_0+a_1+a_2+a_3=a_0\frac{1-r^4}{1-r}=\frac{1}{240}$$
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

 

[Mark44]

not quite sure on what they mean by exact fractions

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

$\dfrac {a}{243}=\dfrac{a(1-r^3)}{240}$

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.

Does that mean my approach is wrong?

Yes, see above.

 

[chwala]

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]

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 $r_2=-\dfrac{1}{3}$

 

[Mark44]

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

Be sure to check these, if you haven't done so already.