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.