Question

I have an arithmetic sequence Question with the picture included

Answer 1

We have the next equation for arithmetics sequences:

`A_n=A_1+(n-1)d`

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:

`-11=A_1+(4-1)d`

and, the 19th term is 34. Then:

`34=A_1+(19-1)d`

So far, we have two variables and two equations.

Hence, we need to solve the system of equations:

Equation 1:

`\begin{gathered} -11=A_(1)+(4-1)d \\ -11=A_1+3d \end{gathered}`

Equation 2:

`\begin{gathered} 34=A_(1)+(19-1)d \\ 34=A_1+18d \end{gathered}`

Now, we can subtract both equations:

`\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=-(45)/(15) \\ d=3 \end{gathered}`

To find A1, we can use any equation to replace the d value:

`\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}`

In conclusion:

The first term is -20.

The common difference is 3.