Program to find Nth term of series 7, 21, 49, 91, 147, 217, ……
Given a number N. The task is to write a program to find the Nth term of the below series:
7, 21, 49, 91, 147, 217, 301, 399, …(N Terms)
Examples:
Input: N = 4
Output: 91
For N = 4
4th Term = ( 7 * 4 * 4 - 7 * 4 + 7)
= 91
Input: N = 10
Output: 636
Given series is:
7, 21, 49, 91, 147, 217, 301, 399, …..
On taking 7 commons from all of the terms, we get:
7 * (1, 3, 7, 13, 21, 31,…..), …..
Now, for the inner-series: 1,3,7,13,21,…
On careful observation we can express the terms of above series as:
1 = (12) – (1-1)
3 = (22) – (2-1)
7 = (32) – (3-1)
13 = (42) – (4-1)
21 = (52) – (5-1)
.
.
.
n-th term = (n2) – (n-1)
Therefore, the n-th term of the actual series will be:
N-th term = 7 * ((n2) - (n-1))
= 7 * (n2 - n + 1)
Below is the implementation of the above approach:
C++
#include <iostream>
#include <math.h>
using namespace std;
int nthTerm( int n)
{
return 7 * pow (n, 2) - 7 * n + 7;
}
int main()
{
int N = 4;
cout << nthTerm(N);
return 0;
}
|
Java
import java.util.*;
class solution
{
static int nthTerm( int n)
{
return 7 * ( int )Math.pow(n, 2 ) - 7 * n + 7 ;
}
public static void main(String arr[])
{
int N = 4 ;
System.out.println(nthTerm(N));
}
}
|
Python3
def nthTerm( n):
return 7 * pow (n, 2 ) - 7 * n + 7
N = 4
print (nthTerm(N))
|
C#
using System;
class GFG
{
static int nthTerm( int n)
{
return 7 * ( int )Math.Pow(n, 2) - 7 * n + 7;
}
public static void Main()
{
int N = 4;
Console.WriteLine(nthTerm(N));
}
}
|
PHP
<?php
function Sum_upto_nth_Term( $n )
{
$r = 7 * pow( $n , 2) - 7 * $n + 7;
echo $r ;
}
$N = 4;
Sum_upto_nth_Term( $N );
?>
|
Javascript
<script>
function nthTerm(n)
{
return 7 * Math.pow(n, 2) - 7 * n + 7;
}
let N = 4;
document.write(nthTerm(N));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
07 Aug, 2022
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