Given a positive integer N, the task is to find the sum of the first N terms of the series
2, 5, 8, 11, 14..
Examples:
Input: N = 5
Output: 40Input: N = 10
Output: 155
Approach:
1st term = 2
2nd term = (2 + 3) = 5
3rd term = (5 + 3) = 8
4th term = (8 + 3) = 11
.
.
Nth term = (2 + (N – 1) * 3) = 3N – 1
The sequence is formed by using the following pattern. For any value N-
Here,
a is the first term
d is the common difference
The above solution can be derived following the series of steps-
The series-
2, 5, 8, 11, …., till N terms
is in A.P. with first term of the series a = 2 and common difference d = 3
Sum of N terms of an A.P. is
Illustration:
Input: N = 5
Output: 40
Explanation:
a = 2
d = 3
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <iostream> using namespace std;
// Function to return // Nth term of the series int nth( int n, int a1, int d)
{ return (a1 + (n - 1) * d);
} // Function to return sum of // Nth term of the series int sum( int a1, int nterm, int n)
{ return n * (a1 + nterm) / 2;
} // Driver code int main()
{ // Value of N
int N = 5;
// First term
int a = 2;
// Common difference
int d = 3;
// finding last term
int nterm = nth(N, a, d);
cout << sum(a, nterm, N) << endl;
return 0;
} |
// C program to implement // the above approach #include <stdio.h> // Function to return // Nth term of the series int nth( int n, int a1, int d)
{ return (a1 + (n - 1) * d);
} // Function to return // sum of Nth term of the series int sum( int a1, int nterm, int n)
{ return n * (a1 + nterm) / 2;
} // Driver code int main()
{ // Value of N
int N = 5;
// First term
int a = 2;
// Common difference
int d = 3;
// Finding last term
int nterm = nth(N, a, d);
printf ( "%d" , sum(a, nterm, N));
return 0;
} |
// Java program to implement // the above approach import java.io.*;
class GFG {
// Driver code
public static void main(String[] args)
{
// Value of N
int N = 5 ;
// First term
int a = 2 ;
// Common difference
int d = 3 ;
// Finding Nth term
int nterm = nth(N, a, d);
System.out.println(sum(a, nterm, N));
}
// Function to return
// Nth term of the series
public static int nth( int n,
int a1, int d)
{
return (a1 + (n - 1 ) * d);
}
// Function to return
// sum of Nth term of the series
public static int sum( int a1,
int nterm, int n)
{
return n * (a1 + nterm) / 2 ;
}
} |
# Python code for the above approach # Function to return # Nth term of the series def nth(n, a1, d):
return (a1 + (n - 1 ) * d);
# Function to return sum of # Nth term of the series def sum (a1, nterm, n):
return n * (a1 + nterm) / 2 ;
# Driver code # Value of N N = 5 ;
# First term a = 2 ;
# Common difference d = 3 ;
# finding last term nterm = nth(N, a, d);
print (( int )( sum (a, nterm, N)))
# This code is contributed by gfgking |
using System;
public class GFG
{ // Function to return
// Nth term of the series
public static int nth( int n, int a1, int d)
{
return (a1 + (n - 1) * d);
}
// Function to return
// sum of Nth term of the series
public static int sum( int a1, int nterm, int n)
{
return n * (a1 + nterm) / 2;
}
static public void Main()
{
// Code
// Value of N
int N = 5;
// First term
int a = 2;
// Common difference
int d = 3;
// Finding Nth term
int nterm = nth(N, a, d);
Console.Write(sum(a, nterm, N));
}
} // This code is contributed by Potta Lokesh |
<script> // JavaScript code for the above approach
// Function to return
// Nth term of the series
function nth(n, a1, d) {
return (a1 + (n - 1) * d);
}
// Function to return sum of
// Nth term of the series
function sum(a1, nterm, n) {
return n * (a1 + nterm) / 2;
}
// Driver code
// Value of N
let N = 5;
// First term
let a = 2;
// Common difference
let d = 3;
// finding last term
let nterm = nth(N, a, d);
document.write(sum(a, nterm, N) + '<br>' );
// This code is contributed by Potta Lokesh
</script>
|
Output:
40
Time complexity: O(1) because performing constant operations
Auxiliary Space: O(1) // since no extra array or recursion is used so the space taken by the algorithm is constant