Help with sequences

[-Whiplash-]

I did a pre-calculus assignment and I recently got it back, I was told to do it over again because I failed it because I did not show my work.

I'm doing pre-calculus by correspondence. (which I hate BTW, since I have no one to ask for help)

Anyway....

the question is as follows:
t3+t4=36
t5+t6=144 find all geometric sequences in which this is true.
Now, originally I did this but didn't include much of my work on the page which got my a 0/10 on the question. but what I did was:

I factored 32 and 144 to find the common factors of 3 and 2, then I applied the geometric formula to find that 3x2^n-1 gives me a correct formula, and results in:
t1=3
t2=6
t3=12
t4=24
t5=48
t5=96
since 12+24 = 36
and 48+96=144 I figured that would be good enough but apparently not so. what should I do? how would I go about solving this? the marker wrote under the question that I should "look for the common ratio" but I'm not sure if he means factor it out like I did or not, is there something I'm missing?

 

[stapel]

t3+t4=36
t5+t6=144 find all geometric sequences in which this is true.

I don't understand what you're doing, so I won't comment on that. Instead, I would suggest using the properties of geometric sequences.

The first term you're working with, t₃, is some number. Then, by nature of geometric sequences, t₄ = r*t₃, so t₃ + t₄ = t₃ + r*t₃ = t₃(1 + r) = 36. In the same way, t₅ = r*t₄ = r*r*t₃ = r²*t₃. What then is t₆?

Factor the expression for t₅ + t₆. Substitute from the equation for t₃ + t₄. What does this give you for the value of r²? Where does this lead?

 

[pka]

the question is as follows:
t3+t4=36
t5+t6=144 find all geometric sequences in which this is true.

First, read the question very carefully. The wording suggests there is more that one answer.
If \(\displaystyle a\) is the first term and \(\displaystyle r\) is the common ratio then \(\displaystyle t_n=a\cdot r^{n-1}\).
From the given we get
\(\displaystyle \\ar^2+ar^3=36\\ar^4+ar^5=144\).

Now you can solve for \(\displaystyle a~\&~r\).

 

[-Whiplash-]

I don't understand what you're doing, so I won't comment on that. Instead, I would suggest using the properties of geometric sequences.

The first term you're working with, t₃, is some number. Then, by nature of geometric sequences, t₄ = r*t₃, so t₃ + t₄ = t₃ + r*t₃ = t₃(1 + r) = 36. In the same way, t₅ = r*t₄ = r*r*t₃ = r²*t₃. What then is t₆?

Factor the expression for t₅ + t₆. Substitute from the equation for t₃ + t₄. What does this give you for the value of r²? Where does this lead?

So thanks to you I figured out this:

T3(1+r)=36/t3r²(1+r)=144

= r²=4
r=2
then I did 36/3=t3
t3=12
so..
12/2 = t2 =6
6/2= t1= 3
thus
t1=3
t2=6
t3=12
t4=24
t5=48
t5=96

But the question says, (as stated before) sequences So I'm not sure what to do now?

 

[pka]

So thanks to you I figured out this:

T3(1+r)=36/t3r²(1+r)=144

= r²=4
r=2
then I did 36/3=t3
t3=12
so..
12/2 = t2 =6
6/2= t1= 3
thus
t1=3
t2=6
t3=12
t4=24
t5=48
t5=96

As I said above, there is more than one answer: \(\displaystyle r=\pm 2\)
So there two values of \(\displaystyle t_1\).

 

[-Whiplash-]

As I said above, there is more than one answer: \(\displaystyle r=\pm 2\)
So there two values of \(\displaystyle t_1\).

but wouldn't -2 make all my answers negative?

 

[pka]

but wouldn't -2 make all my answers negative?

NO INDEED.
If \(\displaystyle r=-2\) then \(\displaystyle t_1=-9\)

\(\displaystyle t_5+t_6=(-9)(-2)^4+(-9)(-2)^5=144\)

Have a look at this webpage. Be sure to click the see more tab.