Find the Nth term in series 12, 35, 81, 173, 357, …
Given a number N, the task is to find the Nth term in series 12, 35, 81, 173, 357, …
Example:
Input: N = 2
Output: 35
2nd term = (12*2) + 11
= 35
Input: N = 5
Output: 357
5th term = (12*(2^4))+11*((2^4)-1)
= 357
Approach:
- Each and every number is obtained by multiplying the previous number by 2 and the addition of 11 to it.
- Since starting number is 12.
1st term = 12
2nd term = (12 * 2) / 11 = 35
3rd term = (35 * 2) / 11 = 81
4th term = (81 * 2) / 11 = 173
And, so on....
-
- In general, Nth number is obtained by formula:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int nthTerm( int N)
{
int nth = 0, first_term = 12;
nth = (first_term * ( pow (2, N - 1)))
+ 11 * (( pow (2, N - 1)) - 1);
return nth;
}
int main()
{
int N = 5;
cout << nthTerm(N) << endl;
return 0;
}
|
Java
class GFG
{
static int nthTerm( int N)
{
int nth = 0 , first_term = 12 ;
nth = ( int ) ((first_term * (Math.pow( 2 , N - 1 )))
+ 11 * ((Math.pow( 2 , N - 1 )) - 1 ));
return nth;
}
public static void main(String[] args)
{
int N = 5 ;
System.out.print(nthTerm(N) + "\n" );
}
}
|
Python3
def nthTerm(N) :
nth = 0 ; first_term = 12 ;
nth = (first_term * ( pow ( 2 , N - 1 ))) + \
11 * (( pow ( 2 , N - 1 )) - 1 );
return nth;
if __name__ = = "__main__" :
N = 5 ;
print (nthTerm(N)) ;
|
C#
using System;
class GFG
{
static int nthTerm( int N)
{
int nth = 0, first_term = 12;
nth = ( int ) ((first_term * (Math.Pow(2, N - 1)))
+ 11 * ((Math.Pow(2, N - 1)) - 1));
return nth;
}
public static void Main(String[] args)
{
int N = 5;
Console.Write(nthTerm(N) + "\n" );
}
}
|
Javascript
<script>
function nthTerm(N)
{
let nth = 0, first_term = 12;
nth = (first_term * (Math.pow(2, N - 1)))
+ 11 * ((Math.pow(2, N - 1)) - 1);
return nth;
}
let N = 5;
document.write(nthTerm(N));
</script>
|
Time complexity: O(log N) for given input N because using inbuilt pow function
Auxiliary Space: O(1)
Last Updated :
28 Jul, 2022
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