I have an arithmetic sequence Question with the picture included

We have the next equation for arithmetics sequences:
[tex]A_n=A_1+(n-1)d[/tex]Where An is the value number of the term given.
A1 is the first term
n is the term number
d is the common difference
If the 4th term is -11. Then:
[tex]-11=A_1+(4-1)d[/tex]and, the 19th term is 34. Then:
[tex]34=A_1+(19-1)d[/tex]So far, we have two variables and two equations.
Hence, we need to solve the system of equations:
Equation 1:
[tex]\begin{gathered} -11=A_{1}+(4-1)d \\ -11=A_1+3d \end{gathered}[/tex]Equation 2:
[tex]\begin{gathered} 34=A_{1}+(19-1)d \\ 34=A_1+18d \end{gathered}[/tex]Now, we can subtract both equations:
[tex]\begin{gathered} -11=A_{1}+3d \\ 34=A_{1}+18d \\ ----------- \\ (-11-34)=(A_1-A_1)+(3d-18d) \\ -45=0-15d \\ Solve\text{ for d} \\ d=-\frac{45}{15} \\ d=3 \end{gathered}[/tex]To find A1, we can use any equation to replace the d value:
[tex]\begin{gathered} 34=A_{1}+18d \\ Where\text{ d=3} \\ 34=A_1+18(3) \\ 34=A_1+54 \\ Solve\text{ for A}_1: \\ A_1=34-54 \\ A_1=-20 \end{gathered}[/tex]In conclusion:
The first term is -20.
The common difference is 3.