**QUESTION**

Find the missing element from an ordered array A[ ], consisting of N elements representing an Arithmetic Progression (AP) .\n \n\nInput:\nThe first line of input contains an integer T denoting the number of test cases. Then T test cases follow. \nThe first line of each test case contains an integer N, where N is the size of the array A[ ].\nThe second line of each test case contains N space separated integers of an Arithmetic Progression denoting elements of the array A[ ].\n \n\nNote: The series should have a missing element in between a perfect A.P. with no missing element will not be considered.\n \n\nOutput:\nPrint out the missing element. \n \n\nConstraints:\n1 <= T <= 100\n2 <= N <= 10\n-50 <= A[i] <=50\n

**ANSWER**

```
#include<stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,i;
scanf("%d",&n);
int a[n];
for(i=0;i<n;i++)
scanf("%d",&a[i]);
if(n==2)
printf("%d",(a[0]+a[1])/2);
int d;
if(n==3)
{
if((a[1]-a[0])<(a[2]-a[1]))
{
d=a[1]-a[0];
printf("%d",a[1]+d);
}
else
{
d=a[2]-a[1];
printf("%d",a[0]+d);
}
}
else
{
if((a[1]-a[0])<(a[2]-a[1]))
d=a[1]-a[0];
else
d=a[2]-a[1];
for(i=1;i<n;i++)
{
if((a[i]-a[i-1])!=d){
printf("%d",a[i-1]+d);
break;
}
}
}
printf("\n");
}
}
```