Please help simplify and reduce the program

Using Arduino Programming Questions
1

Hi, someone is helping me. It's a program that I wrote, and it's slower. The way is the program. If you can help me

word index = 0;
float D1=0;
float D2=0;
float D3=0;
float D4=0;
float D5=0;
.
.
.
float D100=0;

void setup() {
Serial.begin(9600);  
}

void loop() {
 index = index + 1;
 delay(100);
 
 if (index==1)){
 D1= analogRead(A0)/23;
}
 if (index==2)){
 D2= analogRead(A0)/23;
}
 if (index==3)){
 D3= analogRead(A0)/23;
}
 if (index==4)){
 D4= analogRead(A0)/23;
}
 if (index==5)){
 D5= analogRead(A0)/23;
}
.
.
.
 if (index==100)){
 D100= analogRead(A0)/23;
}

if (index>100)){
Serial.println(D1); 
Serial.println(D2);
Serial.println(D3);
Serial.println(D4); 
.
.
.
Serial.println(D100); 
}
}
2 
void loop() {
 index = index + 1;

 
 if (index==1)){
 D1= analogRead(A0)/23;
}

You're welcome

3

What method should I use to reduce the size of the program?

4

Smaller source, smaller compiled code or smaller RAM?

5

I do not want to lose much RAM. I want to reduce the length of the program. What loop should I use?

6

Your program uses a lot of float variables, which isn't necessary as analogRead() outputs an integer from 0-1023, so only 1024 possible values, and using division also is slow. Consider rewriting the math so you handle the read in a different way that keeps everything as integers until the final print formatting.

7

I want to reduce the number of program lines. What should I do?

8

Use arrays and for loops.
Probably.

9

What should I do if I want to define at first the definition of D from 1 to 100 floats?

10

ali_elect62:
What should I do if I want to define at first the definition of D from 1 to 100 floats?

Use an array