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.
On
Let $S_L$ be the sum of numbers before the chosen number and $S_R$ sum of numbers after. Let $D = S_R-S_L$
To begin with $D=13+15+\cdots+59=30^2-6^2$
When we move to the next number we see that the difference $D$ decreases by $11+13=24$ and in the following step a further $13+15=28$.
Hence after $n$ moves, the difference is $D-[24+28+...+(4n+20)]= D-\dfrac{n}{2} (44+4n)=D-2n(n+11)$
We need to find $n$ such that $D-2n(n+11)=0$ i.e. $2n(n+11) = 30^2-6^2 =24\times 36$
or $n(n+11)=12 \times 36 = 16 \times 27$. We can see that $n=16$ at which stage we are at the number $i=43$
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 $$\begin{align} S'_j&=S_{i-1}\\ j(60-j)&=(i-1)(10+\overline{i-1})\\ (25-i)(35+i)&=(i-1)(i+9)\\ i^2+9i-442&=0\\ (i-17)(i+26)&=0\\ i&=17 \quad \color{lightgray}{(i>0)}\qquad\blacksquare\end{align}$$
*The general expression for the last $j$ terms in an AP (derived using the method outlined above) is given by $$\begin{align} S'_j &=\frac j2 [2a+(2n-j-1)d]\\ &=\left[a+(n-\frac12)d\right]j-\frac d2 j^2\\ &=\left(L+\frac d2\right)j-\frac d2j^2 \end{align}$$