Given three integers ‘A’ denotes the primary time period of an arithmetic sequence, ‘C’ denotes the widespread distinction between an arithmetic sequence and an integer ‘B’, the duty is to inform whether or not ‘B’ exists within the arithmetic sequence or not. Return 1 if B is current within the sequence. In any other case, returns 0.
Examples:
Enter: A = 1, B = 3, C = 2
Output: 1
Rationalization: 3 is the second time period of the sequence beginning with 1 and having a typical distinction 2.Enter: A = 1, B = 2, C = 3
Output: 0
Rationalization: 2 will not be current within the sequence.
Method: To unravel the issue observe the under thought:
To unravel this drawback you should utilize the idea of Arithmetic Development. That’s nth time period of an Arithmetic sequence beginning with ‘a’ and customary distinction ‘d’ is a + (n1)*d.
Observations:
 The primary time period is A. If B is bigger than A, then C should be optimistic.
 If B is lesser than A, then C should be damaging.
 Suppose d = distinction between A and B. If B desires to exist within the arithmetic sequence, d should be divisible by C.
Under are the steps for the above method:
 Initialize a variable say, d that may retailer the distinction between B and A.
 Test if B is the same as A, then the time period b is legitimate,
 Test if d < 0, then the widespread distinction should be damaging and d should be divisible by the widespread distinction, if each are usually not the case, return 0.
 if(C ≥ 0), return 0.
 if(dpercentC == 0), return 1.
 If d is bigger than or equal to 0, then the widespread distinction should be better than 0 and d should be divisible by the widespread distinction, if each will not be the case, return 0.
 if(C ≤ 0), return 0.
 if(dpercentC == 0), return 1.
 Else, return 0.
Under is the code for the above method:
C++

Time Complexity: O(1)
Auxiliary House: O(1)