I'm not even sure if this is considered algebra to be honest.. but I'm guessing it is. I'm missing the same notes as some of my friends of course so I'm completely lost as to what to do for a few of these...

If I were to calculate the formula of this arithmetic sequence:
{-1, -3, -5, -7,...}
d1 is 2 right, and t1 (term one) is -1 so the formula would be, uhh, tn=t1+d1(n-1), or tn=-1+2(n-1) or tn=-1+2n-2... is that right?

The second one is if tn=-2n^2-3, find the first 5 terms. The first term is -2 right? but what do I do after that?

The third one... if tn=n^2-11, find which term has a value of 89... I think I go 89=n^2-11 but I'm still confused and not sure if that's right ?

Also, the hardest one that completely confuses me is Determine the equation of this quadratic sequence {3, 4, -1, -12, -29,...} Any advice on how to get that started?

Some help in working out these problems (and therefore problems similar to these) would be greatly appreciated

2. Originally Posted by ~NeonFire372~
If I were to calculate the formula of this arithmetic sequence: {-1, -3, -5, -7,...} d1 is 2 right, and t1 (term one) is -1
I will guess that "d1" is the common difference. If so, then d1 = -2, because the values are getting smaller.

Originally Posted by ~NeonFire372~
The second one is if tn=-2n^2-3, find the first 5 terms. The first term is -2 right? but what do I do after that?
When n = 1, then an = -2(1)^2 - 3 = -2 - 3 = -5. To find the next few terms, plug in 2 for n, then 3 for n, etc, etc.

Originally Posted by ~NeonFire372~
The third one... if tn=n^2-11, find which term has a value of 89... I think I go 89=n^2-11 but I'm still confused and not sure if that's right ?
Do you know how to solve linear equations? (Yes, this is quadratic, but the first step would be one that you would have learned for linear equations, and the answer would be fairly obvious after that one step....)

Originally Posted by ~NeonFire372~
Determine the equation of this quadratic sequence {3, 4, -1, -12, -29,...} Any advice on how to get that started?
Do you know how to solve systems of linear equations? Because you can plug n = 1, n = 2, and n = 3 into the generic quadratic equation, plug in the output values, and thus create a system of linear equations where you'd solve for the coefficients of an^2 + bn + c:

a + b + c = 3
4a + 2b + c = 4
9a + 3b + c = -1

3. Originally Posted by ~NeonFire372~
I'm not even sure if this is considered algebra to be honest.. but I'm guessing it is. I'm missing the same notes as some of my friends of course so I'm completely lost as to what to do for a few of these...

If I were to calculate the formula of this arithmetic sequence:
{-1, -3, -5, -7,...}
d1 is 2 right, and t1 (term one) is -1 so the formula would be, uhh, tn=t1+d1(n-1), or tn=-1+2(n-1) or tn=-1+2n-2... is that right?
If d1 is the common difference than it's -2 since the sequence gets smaller.

The second one is if tn=-2n^2-3, find the first 5 terms. The first term is -2 right? but what do I do after that?
$t_n=2n^2-3$. To find each term do f(1) to f(5).

The third one... if tn=n^2-11, find which term has a value of 89... I think I go 89=n^2-11 but I'm still confused and not sure if that's right ?
Yes. Recall the difference of two squares to solve this.

Also, the hardest one that completely confuses me is Determine the equation of this quadratic sequence {3, 4, -1, -12, -29,...} Any advice on how to get that started?

Some help in working out these problems (and therefore problems similar to these) would be greatly appreciated
$f(x) = ax^2+bx+c \: , \: a \neq 0$
$f(1) = a+b+c = 3$
$f(2) = 16a + 4b + c = 4$
$f(3) = 9a +3b + c = -1$