[tex]\bf \textit{sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n}{2}(a_1+a_n)\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
a_n=\textit{value of the }n^{th}\ term\\
----------\\
n=43\\
a_1=-13\\
S_n=9374
\end{cases}
\\\\\\
9374=\cfrac{43}{2}(-13+a_{43})\implies 18748=43(-13+a_{43})
\\\\\\
\cfrac{18748}{43}=-13+a_{43}\implies 436+13=a_{43}\implies \boxed{449=a_{43}}[/tex]
so.. .we know the first term is -13, and the 43rd term's value is 449...so...this is an arithmetic sequence, thus, in order for us to get from -13 to 449 in 43 hops, we had to add "d" the common difference, 42 times.
so -13, -13 + d, (-13+d)+d, (-13+d+d)+d, ..... and so on.
but in short d + d + d + ... 42 times is just 42d
so, we know that
[tex]\bf \begin{array}{llll}
-13+42d=&449\\
\ \uparrow &\ \uparrow\\
a_1&a_{43}
\end{array}\implies 42d=462\implies d=\cfrac{462}{42}\implies \boxed{d=11}[/tex]
