In the table, the numbers in each row form an arithmetic sequence when read from left to right. Similarly, the numbers in each column form an arithmetic sequence when read from top to bottom. What is the sum of the digits of the value of $x$ ?

(An arithmetic sequence is a sequence in which each term

|  |  |  |  | 18 |
| :--- | :--- | :--- | :--- | :--- |
|  | 43 |  |  |  |
|  |  | 40 |  |  |
|  |  |  |  |  |
| $x$ |  |  | 26 |  |

after the first is obtained from the previous term by adding a constant. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.)
(A) 5
(B) 2
(C) 10
(D) 7
(E) 13
We begin by adding variables to some of the blanks in the grid to make it easier to refer to specific entries:

|  |  |  |  | 18 |
| :---: | :---: | :---: | :---: | :---: |
|  | 43 | $f$ | $g$ | $h$ |
|  |  | 40 |  | $j$ |
|  |  | $k$ |  | $m$ |
| $x$ | $n$ | $p$ | 26 | $q$ |

Since the numbers in each row form an arithmetic sequence and the numbers in each column form an arithmetic sequence, we will refer in several sequences to the common difference.

Let the common difference between adjacent numbers as we move down column 3 be $d$.

Therefore, $k=40+d$ and $p=k+d=40+2 d$.

Also, $40=f+d$, or $f=40-d$.

Moving from left to right along row 2 , the common difference between adjacent numbers must be $(40-d)-43=-3-d$.

Therefore,

$$
\begin{aligned}
& g=f+(-3-d)=(40-d)+(-3-d)=37-2 d \\
& h=g+(-3-d)=(37-2 d)+(-3-d)=34-3 d
\end{aligned}
$$

This gives:

|  |  |  |  | 18 |
| :---: | :---: | :---: | :---: | :---: |
|  | 43 | $40-d$ | $37-2 d$ | $34-3 d$ |
|  |  | 40 |  | $j$ |
|  |  | $40+d$ |  | $m$ |
| $x$ | $n$ | $40+2 d$ | 26 | $q$ |

Moving from the top to the bottom of column 5, the common difference between adjacent numbers can be found by subtracting 18 from $34-3 d$.

That is, the common difference between adjacent numbers in column 5 is $(34-3 d)-18=16-3 d$. Moving down column 5, we can see that

$$
\begin{aligned}
& j=(34-3 d)+(16-3 d)=50-6 d \\
& m=(50-6 d)+(16-3 d)=66-9 d \\
& q=(66-9 d)+(16-3 d)=82-12 d
\end{aligned}
$$

In each case, we have added the common difference to the previous number to obtain the next number.

This gives the following updated grid:

|  |  |  |  | 18 |
| :---: | :---: | :---: | :---: | :---: |
|  | 43 | $40-d$ | $37-2 d$ | $34-3 d$ |
|  |  | 40 |  | $50-6 d$ |
|  |  | $40+d$ |  | $66-9 d$ |
| $x$ | $n$ | $40+2 d$ | 26 | $82-12 d$ |

In row 5 , the difference between the fourth and third entries must equal the difference between the fifth and fourth entries.

In other words,

$$
\begin{aligned}
26-(40+2 d) & =(82-12 d)-26 \\
-14-2 d & =56-12 d \\
10 d & =70 \\
d & =7
\end{aligned}
$$

We can then substitute $d=7$ into our grid to obtain:

|  |  |  |  | 18 |
| :---: | :---: | :---: | :---: | :---: |
|  | 43 | 33 | 23 | 13 |
|  |  | 40 |  | 8 |
|  |  | 47 |  | 3 |
| $x$ | $n$ | 54 | 26 | -2 |

This allows us to determine the value of $x$ by moving along row 5 .

We note that $26-54=-28$ so the common difference in row 5 is -28 .

Therefore, $n+(-28)=54$ or $n=54+28=82$.

Similarly, $x+(-28)=n=82$ and so $x=82+28=110$.

Therefore, the sum of the digits of the value of $x=110$ is $1+1+0=2$.

ANSwER: (B)