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Minimize Nth term of an Arithmetic progression (AP)

  • Last Updated : 14 Apr, 2021

Given two integers A, B which are any two terms of an Arithmetic Progression series, and an integer N, the task is to minimize the Nth term of that arithmetic progression.

Note: All the elements of an AP series must be positive.

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Examples:

Input: A = 1, B = 6, N = 3
Output: 11
Explanations: First three terms of the given arithmetic progression are: {1, 6, 11}.
Therefore, the required output is 11.

Input: A = 20, B = 50, N = 5
Output: 50
Explanations: First Five terms of the given arithmetic progression are: {10, 20, 30, 40, 50}.
Therefore, the required output is 50.



Approach: The problem can be solved by placing A and B at all possible positions in an Arithmetic Progression and check which generates the least possible Nth term. Considering the positions of A and B to be i and j respectively, then the following calculations need to be made:

Common Difference(D) of an AP = (B – A)/(j – i)
First term of an AP = A – (i – 1) * D
Nth term of an AP = First Term + (N – 1)* D 

Finally, return the smallest Nth term obtained.
Follow the steps below to solve the problem:

  1. Initialize a variable, say res to store the smallest possible value of the required Nth term of arithmetic progression.
  2. Using two nested loops, place A and B at all possible positions in an AP, and calculate the Nth term using above calculations. Keep updating res to store the least value of Nth term obtain.
  3. Finally, print the value of res as the required answer.

Below is the implementation of the above approach:
 

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the smallest
// Nth term of an AP possible
int smallestNth(int A, int B, int N)
{
    // Stores the smallest Nth term
    int res = INT_MAX;
 
    for (int i = 1; i < N; i++) {
        for (int j = N; j > i; j--) {
 
            // Check if common difference
            // of AP is an integer
            if ((B - A) % (j - i) == 0) {
 
                // Store the common
                // difference
                int D = (B - A) / (j - i);
 
                // Store the First Term
                // of that AP
                int FirstTerm = A - (i - 1) * D;
 
                // Store the Nth term of
                // that AP
                int NthTerm = FirstTerm + (N - 1) * D;
 
                // Check if all elements of
                // an AP are positive
                if (FirstTerm > 0)
                    res = min(res, NthTerm);
            }
        }
    }
 
    // Return the least
    // Nth term obtained
    return res;
}
 
// Driver Code
int main()
{
    int N = 3;
    int A = 1;
    int B = 6;
    cout << smallestNth(A, B, N);
}

Java




// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
  
// Function to find the smallest
// Nth term of an AP possible
static int smallestNth(int A, int B, int N)
{
 
    // Stores the smallest Nth term
    int res = Integer.MAX_VALUE;
  
    for(int i = 1; i < N; i++)
    {
        for(int j = N; j > i; j--)
        {
             
            // Check if common difference
            // of AP is an integer
            if ((B - A) % (j - i) == 0)
            {
                 
                // Store the common
                // difference
                int D = (B - A) / (j - i);
  
                // Store the First Term
                // of that AP
                int FirstTerm = A - (i - 1) * D;
  
                // Store the Nth term of
                // that AP
                int NthTerm = FirstTerm + (N - 1) * D;
  
                // Check if all elements of
                // an AP are positive
                if (FirstTerm > 0)
                    res = Math.min(res, NthTerm);
            }
        }
    }
  
    // Return the least
    // Nth term obtained
    return res;
}
  
// Driver Code
public static void main (String[] args)
{
    int N = 3;
    int A = 1;
    int B = 6;
  
    System.out.print(smallestNth(A, B, N));
}
}
 
// This code is contributed by code_hunt

Python3




# Python3 program to implement
# the above approach
import sys
 
# Function to find the smallest
# Nth term of an AP possible
def smallestNth(A, B, N):
     
    # Stores the smallest Nth term
    res = sys.maxsize
 
    for i in range(1, N):
        for j in range(N, i, -1):
 
            # Check if common difference
            # of AP is an integer
            if ((B - A) % (j - i) == 0):
 
                # Store the common
                # difference
                D = (B - A) // (j - i)
 
                # Store the First Term
                # of that AP
                FirstTerm = A - (i - 1) * D
 
                # Store the Nth term of
                # that AP
                NthTerm = FirstTerm + (N - 1) * D
 
                # Check if all elements of
                # an AP are positive
                if (FirstTerm > 0):
                    res = min(res, NthTerm)
 
    # Return the least
    # Nth term obtained
    return res
 
# Driver Code
if __name__ == '__main__':
     
    N = 3
    A = 1
    B = 6
     
    print(smallestNth(A, B, N))
     
# This code is contributed by mohit kumar 29

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
  
// Function to find the smallest
// Nth term of an AP possible
static int smallestNth(int A, int B, int N)
{
     
    // Stores the smallest Nth term
    int res = Int32.MaxValue;
  
    for(int i = 1; i < N; i++)
    {
        for(int j = N; j > i; j--)
        {
             
            // Check if common difference
            // of AP is an integer
            if ((B - A) % (j - i) == 0)
            {
                 
                // Store the common
                // difference
                int D = (B - A) / (j - i);
  
                // Store the First Term
                // of that AP
                int FirstTerm = A - (i - 1) * D;
  
                // Store the Nth term of
                // that AP
                int NthTerm = FirstTerm + (N - 1) * D;
  
                // Check if all elements of
                // an AP are positive
                if (FirstTerm > 0)
                    res = Math.Min(res, NthTerm);
            }
        }
    }
  
    // Return the least
    // Nth term obtained
    return res;
}
  
// Driver Code
public static void Main ()
{
    int N = 3;
    int A = 1;
    int B = 6;
     
    Console.Write(smallestNth(A, B, N));
}
}
 
// This code is contributed by code_hunt

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find the smallest
// Nth term of an AP possible
function smallestNth(A, B, N)
{
  
    // Stores the smallest Nth term
    let res = Number.MAX_VALUE;
   
    for(let i = 1; i < N; i++)
    {
        for(let j = N; j > i; j--)
        {
              
            // Check if common difference
            // of AP is an integer
            if ((B - A) % (j - i) == 0)
            {
                  
                // Store the common
                // difference
                let D = (B - A) / (j - i);
   
                // Store the First Term
                // of that AP
                let FirstTerm = A - (i - 1) * D;
   
                // Store the Nth term of
                // that AP
                let NthTerm = FirstTerm + (N - 1) * D;
   
                // Check if all elements of
                // an AP are positive
                if (FirstTerm > 0)
                    res = Math.min(res, NthTerm);
            }
        }
    }
   
    // Return the least
    // Nth term obtained
    return res;
}
 
// Driver code
    let N = 3;
    let A = 1;
    let B = 6;
   
    document.write(smallestNth(A, B, N));
 
// This code is contributed by target_2.
</script>
Output: 
11

 

Time Complexity: O(N2)
Auxiliary Space: O(1)




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