What is the $46^{th}$ term in the sequence $27,-20,13,-6,-1,8,-15$ I am doing some practice algebra from brilliant.org and I am having some trouble with the following question:


What is the $46^{th}$ term in the sequence $27,-20,13,-6,-1,8,-15$


I tried finding the common difference and I also tried to ignoring the minus sign and came up with the following pattern of the differences or sums:
$$-7,-7,-7,5,7,7,7$$
but that didn't help. If someone could provide some hints, that would be awesome.
Thanks!
 A: Hint: Split the sequence into two:
$$a_{2n+1}=27-14n$$
$$a_{2n}=-20+14n$$
A: $27+-20=7\\-20+13=-7\\13+-6=7\\-6+-1=-7\\-1+8=7\\\vdots$
$a_n+a_{n-1}=(-1)^n\times7$
$${ a }_{ 1 }+{ a }_{ 2 }=7\\ -{ a }_{ 2 }-{ a }_{ 3 }=7\\ { a }_{ 3 }+{ a }_{ 4 }=7\\ \vdots \\ -{ a }_{ 44 }-{ a }_{ 45 }=7\\ ---------\\ { a }_{ 1 }-{ a }_{ 45 }=7\times 44\\ { a }_{ 45 }=27-7\times 44=-281\\ ---------\\ { a }_{ 45 }+{ a }_{ 46 }=7\\ { a }_{ 46 }=288$$
If my calculations are correct the answer is $288$
A: It is impossible to tell. 
For example, there are infinitely many sequences which start $2,4,\ldots$
For any $k \in \mathbb{C}$, the sequence $s_n :=kn^2+(2-3k)n+2k$ starts $s_1=2$ and $s_2=4$. 
Moreover, you may choose $k$ to make $s_p$, with $p \ge 3$, to be anything you like.
The same is true for your sequence. There are an uncountably infinitely number of sequence of degree seven in $n$ for which their first seven terms are the same as the ones you have given. 
A: The numbers given so far satisfy the recurrence relation:
$$a_{n+2}+2 a_{n+1} + a_n = 0$$
Since the associated equation $\lambda^2 + 2\lambda + 1 = (\lambda+1)^2$ has a double root at $-1$, the recurrence relation has general solution of the form
$$a_n = (-1)^n (\alpha n + \beta)$$
Substituting the initial condition $a_1 = 27, a_2 = -20$ into above formula give us $\alpha = 7, \beta = -34$. The corresponding formula $\displaystyle a_n = (-1)^{n-1}(34-7n)$ does reproduce the initial sequences. The $46^{th}$ term will be $(-1)^{46-1}(34-7\cdot46) = 288$.
A: The total delta between each term changes by 14 each time.


*

*$|-20-27| = 47$

*$|13-(-20)| = 33$

*$|-6-13| = 19$

*$|-1-(-6)| = 5$

*Now when we subtract 14 from 5, we get -9, but we're just looking at the magnitude for now...

*$|8-(-1)| = 9$

*$|-15-8| = 23$


Now, appropriately add $\pm$ signs.
