An arithmetic progression is defined by two numbers: the value of the first term, $A_0,$ and the common

difference, $D.$ Each term can be computed from the previous term by adding the common difference

$D$ to it$.$ In other words,

$A_i=A_{i-1}+D,i=1,2,$dots

Using the above formula, we can find the $n^{th}$ term of the progression as follows:

$A_1=A_0+D$

$A_2=A_1+D$

$A_3=A_2+D$

cdots

$A_i=A_{i-1}+D$

cdots

$A_n=A_{n-1}+D$

For example, if the first term $A_0=3,$ and the common difference $D=4,$ we compute the $6^{th}$ term of

the progression, $A_5$ as follows:

$A_1=3+4=7$

$A_2=7+4=11$

$A_3=11+4=15$

$A_4=15+4=19$

$A_5=19+4=23$

Write an assembly program that computes the $n^{th}$ term of the arithmetic progression. The program

declares the following identifiers of type SDWORD:

$A0$ : the value of the first term,

$D$ : the common difference,

$N$ : the rank of the term of the progression to be computed.

When the program completes, the value of the $n^{th}$ term, $A_n,$ shall be in register EAX. Use the

following cases to test your program:

$A_0=1,D=1,N=9,$ the program computes $A_9=10$inEAx.

$A_0=3,D=4,N=5,$ the program computes $A_5=23$inEAx.

$A_0=85,D=-7,N=7,$ the program computes $A_7=36in$ EAx.