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Find Nth term of the series where each term differs by 6 and 2 alternately

  • Last Updated : 15 Mar, 2021

Given a number N, the task is to find the Nth term of the series where each term differs by 6 and 2 alternately.
Examples: 
 

Input: N = 6 
Output: 24 
Explanation: 
The Nth term is 0 + 6 + 2 + 6 + 2 + 6 + 2 = 24
Input: N = 3 
Output: 14 
Explanation: 
The Nth term is 0 + 6 + 2 + 6 = 14 
 

 

Naive Approach: The idea is to iterate from 1 with an increment of 6 and 2 alternatively, till we reach the Nth term.
Below is the implementation of the above approach: 
 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find Nth term
void findNthTerm(int N)
{
    int ans = 0;
 
    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {
 
        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }
 
        // Else add 2
        else {
            ans = ans + 2;
        }
    }
 
    // Print ans
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    int N = 3;
    findNthTerm(N);
    return 0;
}

Java




// Java program for the above approach
class GFG{
  
// Function to find Nth term
static void findNthTerm(int N)
{
    int ans = 0;
  
    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {
  
        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }
  
        // Else add 2
        else {
            ans = ans + 2;
        }
    }
  
    // Print ans
    System.out.print(ans +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program for the above approach
 
# Function to find Nth term
def findNthTerm(N):
    ans = 0
 
    # Iterate from 1 till Nth term
    for i in range(N):
 
        # Check if i is even and
        # then add 6
        if (i % 2 == 0) :
            ans = ans + 6
         
 
        # Else add 2
        else :
            ans = ans + 2
         
    # Print ans
    print(ans)
 
 
# Driver Code
if __name__=='__main__':
 
    N = 3
    findNthTerm(N)
     
 
# This code is contributed by AbhiThakur

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find Nth term
static void findNthTerm(int N)
{
    int ans = 0;
 
    // Iterate from 1 till Nth term
    for (int i = 0; i < N; i++) {
 
        // Check if i is even and
        // then add 6
        if (i % 2 == 0) {
            ans = ans + 6;
        }
 
        // Else add 2
        else {
            ans = ans + 2;
        }
    }
 
    // Print ans
    Console.Write(ans +"\n");
}
 
// Driver Code
public static void Main()
{
    int N = 3;
    findNthTerm(N);
}
}
 
// This code is contributed by AbhiThakur

Javascript




<script>
 
// JavaScript program for the above approach
   
    // Function to find Nth term
    function findNthTerm(N)
    {
        let ans = 0;
   
        // Iterate from 1 till Nth term
        for (let i = 0; i < N; i++) {
   
            // Check if i is even and
            // then add 6
            if (i % 2 == 0) {
                ans = ans + 6;
            }
   
            // Else add 2
            else {
                ans = ans + 2;
            }
        }
   
        // Print ans
        document.write(ans + "<br>");
    }
   
    // Driver Code
       let N = 3;
       findNthTerm(N);
  
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 



14

 

Time Complexity: O(N)
Efficient Approach: We can find the Nth term by using the below formula: 
 

  1. If N is Odd: The Nth term is given by (N/2 + 1)*6 + (N/2)*2.
  2. If N is Even: The Nth term is given by (N/2)*6 + (N/2)*2.

Below is the implementation of the above approach: 
 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find Nth term
void findNthTerm(int N)
{
    int ans;
 
    // Check if N is even
    if (N % 2 == 0) {
 
        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }
 
    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }
 
    // Print ans
    cout << ans << endl;
}
 
// Driver Code
int main()
{
    int N = 3;
    findNthTerm(N);
    return 0;
}

Java




// Java program for the above approach
 
 
class GFG{
  
// Function to find Nth term
static void findNthTerm(int N)
{
    int ans;
  
    // Check if N is even
    if (N % 2 == 0) {
  
        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }
  
    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }
  
    // Print ans
    System.out.print(ans +"\n");
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}
 
// This code contributed by PrinciRaj1992

Python3




# Python3 program for the above approach
 
# Function to find Nth term
def findNthTerm(N):
    ans = 0;
 
    # Check if N is even
    if (N % 2 == 0):
 
        # Formula for n is even
        ans = (N // 2) * 6 + (N // 2) * 2;
     
    # Check if N is odd
    else:
         
        # Formula for N is odd
        ans = (N // 2 + 1) * 6 + (N // 2) * 2;
     
    # Print ans
    print(ans);
     
# Driver Code
if __name__ == '__main__':
     
    N = 3;
    findNthTerm(N);
     
# This code is contributed by Rajput-Ji

C#




// C# program for the above approach
using System;
 
class GFG{
   
// Function to find Nth term
static void findNthTerm(int N)
{
    int ans;
   
    // Check if N is even
    if (N % 2 == 0) {
   
        // Formula for n is even
        ans = (N / 2) * 6
              + (N / 2) * 2;
    }
   
    // Check if N is odd
    else {
        // Formula for N is odd
        ans = (N / 2 + 1) * 6
              + (N / 2) * 2;
    }
   
    // Print ans
    Console.Write(ans +"\n");
}
   
// Driver Code
public static void Main(String[] args)
{
    int N = 3;
    findNthTerm(N);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to find Nth term
function findNthTerm( N)
{
    let ans;
 
    // Check if N is even
    if (N % 2 == 0) {
 
        // Formula for n is even
        ans = parseInt(N / 2) * 6
              + parseInt(N / 2) * 2;
    }
 
    // Check if N is odd
    else {
        // Formula for N is odd
        ans = parseInt(N / 2 + 1) * 6
              + parseInt(N / 2) * 2;
    }
 
    // Print ans
     document.write(ans);
}
 
// Driver Function
 
    // get the value of N
    let N = 3;
 
    // Calculate and print the Nth term
    findNthTerm(N);
 
// This code is contributed by todaysgaurav
 
</script>
Output: 
14

 

Time Complexity: O(1)
 

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