Arithmetic Sequence - Find the Last Number of Terms

The sequence is: $11, 13, 15, ... 59$

I need to find $i$ so that the sum of all terms before $i$ equals the sum of all terms after $i$

The simple way would be to calculate: $S_{i-1} = S_{n} - S_{i}$

But I would like to solve it a different way. Is there a formula to find the last number of terms of an arithmetic sequence?

I have found something but it doesn't seem to work when I use it. Please also show me how you solve the equation.

Answer

Another method, as requested.

Counting backwards from $59$ using $d'=-2$, the $j$-th term is $T'_j=61-2j$, and the sum of $j$ terms is $S'_j=\frac j2(59+T'_j)=60j-j^2=j(60-j)$, where the dash indicates counting backwards. This is also the formula for the last $j$ terms (counting forwards), as mentioned in your question.*

Counting forwards from $11$, using standard formulas, we have $S_i=10i+i^2=i(10+i)$.

We want the forward sum to $(i-1)$ terms to be equal the backward sum to $j$ terms, and also $i+j=25$, hence
$$\begin{align}
S'_j&=S_{i-1}\\
j(60-j)&=(i-1)(10+\overline{i-1})\\
(25-i)(35+i)&=(i-1)(i+9)\\
i^2+9i-442&=0\\
(i-17)(i+26)&=0\\
i&=17 \quad \color{lightgray}{(i>0)}\qquad\blacksquare\end{align}$$

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*The general expression for the last $j$ terms in an AP (derived using the method outlined above) is given by
$$\begin{align}
S'_j
&=\frac j2 [2a+(2n-j-1)d]\\
&=\left[a+(n-\frac12)d\right]j-\frac d2 j^2\\
&=\left(L+\frac d2\right)j-\frac d2j^2
\end{align}$$