$A+B+C=2149$, Find $A$ In the following form of odd numbers 

If the numbers 
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.

$A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \cdots + (n-1)$ odd numbers before it, which equals $\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\left(\frac{n(n-1)}{2}\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$
This formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.
If we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.
So can we solve:
$$3n^2 + n + 6k - 1 = 2149$$
Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$
Now we need $k$ to be an integer, and $1 \leq k \leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \leq 2150$, so $n^2 \leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!
This gives $A = 26^2 - 26 + 32 - 1 = 681$.
