First we write out our given information:
$$a_8=2a_2$$
$$a_{11}=18$$
$a_n$ is an arithmetic sequence.
Where here $a_n$ means the $n$th term of our sequence.
What does an arithmetic sequence mean? It means to get to the next term in your sequence you add a constant ($c$) each time:
$$a_{n+1}=a_n+c$$
Equivalently:
$$\frac{a_{n+1}-a_{n}}{(n+1)-n}=c$$
So $a_n$ is of slope $c$ ($c_2$ is another constant):
$$a_n=cn+c_2$$
Where here $c_2=a_0$ (Substitute in $n=0$ and see why that has to be the case if we let $a_0$ exist)
Now we use the other given information to try to come up with a solution.
Let $n=2$:
$$a_2=2c+c_2 {}{}$$
Let $n=8$, using the above equation we have:
$$a_8=8c+c_2=2a_2=4c+2c_2 {}{}{}{}$$
Let $n=11$
$$a_{11}=18$$
$$a_{11}=11c+c_2$$
But $a_{11}-a_8=(11c+c_2)-(8c+c_2)=3c$
Hence, $a_{11}=3c+a_{8}$
$$a_{11}=3c+4c+2c_2=18$$
$$a_{11}=3c+8c+c_2=18$$
Solve this system of equations to get a closed form for the arithmetic sequence.
$$a_n=1.2n+4.8$$.
We can check it works $a_2=1.2(2)+4.8=7.2$. Now we compute $a_8$ to see if $a_8=2a_2$ as required: $a_8=1.2(8)+4.8=14.4=2(7.2)=2a_2$. It is arithmetic as we may check $a_{n+1}-a_n$ is a constant $1.2$. Also $a_{11}=1.2(11)+4.8=18$ as required.
The answers follow from this, from summation formulas/methods of evaluating sums, and from algebra.
Best Answer
For an AP, the $n^{th}$ term is given by:
$$a_n=a+(n-1)d$$
,where $a$ is the first term and $d$ is the common difference
In this case
$$a_7=a+6d=-1\tag{1}$$
$$a_{16}=a+15d=17\tag{2}$$
Subtracting $(1)$ from $(2)$, we get
$9d=18 \Rightarrow d=2\tag{3}$
Now, plugging $(3)$ in $(1)$, we get
$$a+12=-1 \Rightarrow a=-13$$
An alternative (but similar) approach:
It is not hard to see that an AP behaves like a linear function.(perhaps from it's definition)
Let the general term of the AP be $$a_n=bn+c$$, where $b$ and $c$ are some constants.
Using this we get
$$a_7=7b+c=-1\tag{1}$$
$$a_{16}=16b+c=17\tag{2}$$
Subtracting $(1)$ from $(2)$, we get
$$9b=18\Rightarrow b=2\tag{3}$$
Using $(3)$ in $(1)$, we get
$$14+c=-1 \Rightarrow c=-15$$
Note:
Geometric version of this is pointed out by Bill Dubuque in a comment.
Using the two approaches we can observe that
$$a+(n-1)d=dn+(a-d)=bn+c$$
Since this an equation in $n$, for it to be an identity the coefficients must be same.
Comparing like power coefficients, we get
$$b=d \text{ and } a-d=c$$