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++ 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 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 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# 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 |
<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> |
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++ 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 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 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# 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 |
<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> |
14
Time Complexity: O(1)
Auxiliary Space: O(1) since using constant variables