$A+B+C=2149$, 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.

If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation.  $\frac{2149}{3}\approx 716$, so $A<716<B<C$.  Which row would $715$ fit into?
By changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\left[\begin{array}{}\color{red}{1}\\2&\color{red}{3}\\4&5&\color{red}{6}\\7&8&9&\color{red}{10}\\\vdots\end{array}\right]$, and these numbers are very familiar to us.  The red numbers are the triangle numbers, $T(n)=\binom{n+1}{2}=\frac{n^2+n}{2}$.
So, $T(26)=351<\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).
So, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$
In the case that $n=25$, this would be $3A+100+2=2149\Rightarrow 3A=2047\Rightarrow A=\frac{2047}{3}\not\in\mathbb{Z}$, so we know that $n=25$ was not possible.
In the case that $n=26$, this would be $3A+104+2=2149\Rightarrow 3A=2043\Rightarrow A=681$
In the case that $n=27$, this would be $3A+108+2=2149\Rightarrow 3A=2039\Rightarrow A=\frac{2039}{3}\not\in\mathbb{Z}$, so we know that $n=27$ was not possible.
We can similarly show that if $n=24$ that leads to a contradiction as well.
We believe then that our answer is $A=681, B=733, C=735$.  All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\frac{A+1}{2}$ and $T(26)$ is greater than $\frac{A+1}{2}$.
Indeed, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$
