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 termvoid 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 Codeint main(){ int N = 3; findNthTerm(N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to find Nth termstatic 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 Codepublic 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 termdef 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 Codeif __name__=='__main__': N = 3 findNthTerm(N) # This code is contributed by AbhiThakur |
C#
// C# program for the above approachusing System;class GFG{// Function to find Nth termstatic 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 Codepublic 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)
Auxiliary Space: O(1) as using constant variables
Efficient Approach: We can find the Nth term by using the below formula:
- If N is Odd: The Nth term is given by (N/2 + 1)*6 + (N/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 termvoid 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 Codeint main(){ int N = 3; findNthTerm(N); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to find Nth termstatic 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 Codepublic 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 termdef 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 Codeif __name__ == '__main__': N = 3; findNthTerm(N); # This code is contributed by Rajput-Ji |
C#
// C# program for the above approachusing System;class GFG{ // Function to find Nth termstatic 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 Codepublic 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 termfunction 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)
Auxiliary Space: O(1) since using constant variables
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