Given a positive integer N, the task is to find the sum of the first N terms of the seriesĀ
2, 5, 8, 11, 14..
Examples:
Input: N = 5
Output: 40Input: N = 10
Output: 155
Ā
Approach:
1st term = 2
2nd term = (2 + 3) = 5
3rd term = (5 + 3) = 8
4th term = (8 + 3) = 11
.
.
Nth term = (2 + (N ā 1) * 3) = 3N ā 1
The sequence is formed by using the following pattern. For any value N-
Here,Ā
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, ā¦., 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 = 2Ā
d = 3
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <iostream>using namespace std;Ā
// Function to return// Nth term of the seriesint nth(int n, int a1, int d){Ā Ā Ā Ā return (a1 + (n - 1) * d);}Ā
// Function to return sum of// Nth term of the seriesint sum(int a1, int nterm, int n){Ā Ā Ā Ā return n * (a1 + nterm) / 2;}Ā
// Driver codeint main(){Ā Ā Ā Ā // Value of NĀ Ā Ā Ā int N = 5;Ā
Ā Ā Ā Ā // First termĀ Ā Ā Ā int a = 2;Ā
Ā Ā Ā Ā // Common differenceĀ Ā Ā Ā int d = 3;Ā
Ā Ā Ā Ā // finding last termĀ Ā Ā Ā int nterm = nth(N, a, d);Ā Ā Ā Ā cout << sum(a, nterm, N) << endl;Ā Ā Ā Ā return 0;} |
C
// C program to implement// the above approach#include <stdio.h>Ā
// Function to return// Nth term of the seriesint nth(int n, int a1, int d){Ā Ā Ā Ā return (a1 + (n - 1) * d);}Ā
// Function to return// sum of Nth term of the seriesint sum(int a1, int nterm, int n){Ā Ā Ā Ā return n * (a1 + nterm) / 2;}Ā
// Driver codeint main(){Ā Ā Ā Ā // Value of NĀ Ā Ā Ā int N = 5;Ā
Ā Ā Ā Ā // First termĀ Ā Ā Ā int a = 2;Ā
Ā Ā Ā Ā // Common differenceĀ Ā Ā Ā int d = 3;Ā
Ā Ā Ā Ā // Finding last termĀ Ā Ā Ā int nterm = nth(N, a, d);Ā
Ā Ā Ā Ā printf("%d", sum(a, nterm, N));Ā Ā Ā Ā return 0;} |
Java
// Java program to implement// the above approachimport java.io.*;Ā
class GFG {Ā Ā Ā Ā // Driver codeĀ Ā Ā Ā public static void main(String[] args)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā // Value of NĀ Ā Ā Ā Ā Ā Ā Ā int N = 5;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // First termĀ Ā Ā Ā Ā Ā Ā Ā int a = 2;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Common differenceĀ Ā Ā Ā Ā Ā Ā Ā int d = 3;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Finding Nth termĀ Ā Ā Ā Ā Ā Ā Ā int nterm = nth(N, a, d);Ā Ā Ā Ā Ā Ā Ā Ā System.out.println(sum(a, nterm, N));Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Function to returnĀ Ā Ā Ā // Nth term of the seriesĀ Ā Ā Ā public static int nth(int n,Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int a1, int d)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā return (a1 + (n - 1) * d);Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Function to returnĀ Ā Ā Ā // sum of Nth term of the seriesĀ Ā Ā Ā public static int sum(int a1,Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int nterm, int n)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā return n * (a1 + nterm) / 2;Ā Ā Ā Ā }} |
Python3
# Python code for the above approach Ā
# Function to return# Nth term of the seriesdef nth(n, a1, d):Ā Ā Ā Ā return (a1 + (n - 1) * d);Ā
# Function to return sum of# Nth term of the seriesdef sum(a1, nterm, n):Ā Ā Ā Ā return n * (a1 + nterm) / 2;Ā
# Driver codeĀ
# Value of NN = 5;Ā
# First terma = 2;Ā
# Common differenced = 3;Ā
# finding last termnterm = nth(N, a, d);print((int)(sum(a, nterm, N)))Ā
# This code is contributed by gfgking |
C#
using System;Ā
public class GFG{Ā Ā Ā Ā Ā Ā Ā // Function to returnĀ Ā Ā Ā // Nth term of the seriesĀ Ā Ā Ā public static int nth(int n, int a1, int d)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā return (a1 + (n - 1) * d);Ā Ā Ā Ā }Ā
Ā Ā Ā Ā // Function to returnĀ Ā Ā Ā // sum of Nth term of the seriesĀ Ā Ā Ā public static int sum(int a1, int nterm, int n)Ā Ā Ā Ā {Ā Ā Ā Ā Ā Ā Ā Ā return n * (a1 + nterm) / 2;Ā Ā Ā Ā }Ā Ā Ā Ā static public void Main()Ā Ā Ā Ā {Ā
Ā Ā Ā Ā Ā Ā Ā Ā // CodeĀ Ā Ā Ā Ā Ā Ā Ā // Value of NĀ Ā Ā Ā Ā Ā Ā Ā int N = 5;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // First termĀ Ā Ā Ā Ā Ā Ā Ā int a = 2;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Common differenceĀ Ā Ā Ā Ā Ā Ā Ā int d = 3;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Finding Nth termĀ Ā Ā Ā Ā Ā Ā Ā int nterm = nth(N, a, d);Ā Ā Ā Ā Ā Ā Ā Ā Console.Write(sum(a, nterm, N));Ā Ā Ā Ā }}Ā
// This code is contributed by Potta Lokesh |
Javascript
<script>Ā Ā Ā Ā Ā Ā Ā Ā // JavaScript code for the above approach Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Function to returnĀ Ā Ā Ā Ā Ā Ā Ā // Nth term of the seriesĀ Ā Ā Ā Ā Ā Ā Ā function nth(n, a1, d) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return (a1 + (n - 1) * d);Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Function to return sum ofĀ Ā Ā Ā Ā Ā Ā Ā // Nth term of the seriesĀ Ā Ā Ā Ā Ā Ā Ā function sum(a1, nterm, n) {Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return n * (a1 + nterm) / 2;Ā Ā Ā Ā Ā Ā Ā Ā }Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Driver codeĀ
Ā Ā Ā Ā Ā Ā Ā Ā // Value of NĀ Ā Ā Ā Ā Ā Ā Ā let N = 5;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // First termĀ Ā Ā Ā Ā Ā Ā Ā let a = 2;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // Common differenceĀ Ā Ā Ā Ā Ā Ā Ā let d = 3;Ā
Ā Ā Ā Ā Ā Ā Ā Ā // finding last termĀ Ā Ā Ā Ā Ā Ā Ā let nterm = nth(N, a, d);Ā Ā Ā Ā Ā Ā Ā Ā document.write(sum(a, nterm, N) + '<br>');Ā
Ā Ā Ā Ā Ā Ā Ā // This code is contributed by Potta LokeshĀ Ā Ā Ā </script> |
Ā
Ā
Output:Ā
40
Ā
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
Ā
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