turbo c find arithmetic progression term a,a+b,..., a+(n+1)d
plizzz help mee((I don`t know C((

Recommended Answers

All 5 Replies

Start learning C first, You can start by reading yashwant or Let US C for learning it. Then you can try Dennis Ricthie for intensive learning. Then write some short codes and then learn what is Arithmetic progression and then try to write this code. You will learn a lot.

please correct my code

#include <stdio.h>
#include<math.h>
int main () {

printf("vvod a,d,n"0;
scanf("%F%f%f",&a,&d,&n);
if (a=a+d*(n-1)){
}
printf("a:%6.2f,d:6.2f,n:%6.2f",a,d,n);
return 0;
}

you didn't declare the variables you used. You have to declare all variables before they can be used, for example:

float a;

If you don't know what "float" is then you should read up on C Language Data Types.

line 7: the = is an assignment operator, the == is the boolean operator. If you want to compare a with something them use ==.

wth are people still trying to use turbo C when that product is going on 35 years old (older than they are almost certainly), hasn't been supported for over 25 of those, and was created for an operating system and hardware architecture they've at most only read about or seen in a museum trip?
Yet if they don't get a new iPhone every few months they start complaining that they have such seriously outdated stuff...

and oh, learn English, learn C, and start by trying to do your own homework.

wth are people still trying to use turbo C

Because many universities are too damned ignorant and lazy to upgrade their skills. And they have the nerve to call themselves institutions of higher learning!!

and oh, learn English

I don't have a problem with bad English grammer from people whose native language is something else. Last time I knew DaniWeb wasn't just for English majors.

commented: Well said. +14
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.