Geeksforgeeks Solution For " Sum of an AP "

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Problem :- Write a program to print sum of an AP with n terms with first term a and common difference d.

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Solution :- 

Method 1 :-

#include <iostream>
using namespace std;
int main() 
{
int t;
cin>>t;
while(t--)
{
   int n,a,b,d,sum=0;
   cin>>n;
   cin>>a>>d;
   sum=(n*((2*a)+((n-1)*d)))/2;
   cout<<sum<<endl;
}
return 0;
}

Method 2 :-

#include <iostream>
#include<cstdio>
#include <cmath>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
       int n,a,p,r;
       long double sum=0.0;
       cin>>n;
       cin>>a>>r;
       if(r==1){
           cout<<a*n<<endl;
           continue;
       }
       p=pow(r,n);
       sum=(a*(1-p)/(1-r));
       sum/=1.0;
       //cout<<sum<<endl;
       printf("%Lf\n", sum);

    }
    return 0;

}

Method 3 :-

#include <stdio.h>
#include <math.h>

double GP(int a, int r, int n) 
{
    return (a*pow(r,n)-a)/(r-1);
}
int main() 
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
      double N, i, a, r;
      scanf("%lf\n",&N);
      scanf("%lf %lf",&a,&r);
      if(r==1)
      {
          printf("%lf\n",a*N);
      }
      else 
      {
      printf("%lf\n", GP(a, r, N));
      }
  }
  return 0;

}

Output:-