Una serie con la misma diferencia común se conoce como serie aritmética . El primer término de la serie es a y la diferencia común es d . La serie se ve como a, a + d, a + 2d, a + 3d, . . . La tarea es encontrar la suma de la serie.
Ejemplos:
Input : a = 1
d = 2
n = 4
Output : 16
1 + 3 + 5 + 7 = 16
Input : a = 2.5
d = 1.5
n = 20
Output : 335
Una solución simple para encontrar la suma de series aritméticas.
C++
// CPP Program to find the sum of arithmetic
// series.
#include<bits/stdc++.h>
using namespace std;
// Function to find sum of series.
float sumOfAP(float a, float d, int n)
{
float sum = 0;
for (int i=0;i<n;i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
// Driver function
int main()
{
int n = 20;
float a = 2.5, d = 1.5;
cout<<sumOfAP(a, d, n);
return 0;
}
Java
// JAVA Program to find the sum of
// arithmetic series.
class GFG{
// Function to find sum of series.
static float sumOfAP(float a, float d,
int n)
{
float sum = 0;
for (int i = 0; i < n; i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
// Driver function
public static void main(String args[])
{
int n = 20;
float a = 2.5f, d = 1.5f;
System.out.println(sumOfAP(a, d, n));
}
}
/*This code is contributed by Nikita Tiwari.*/
Python
# Python Program to find the sum of # arithmetic series. # Function to find sum of series. def sumOfAP( a, d,n) : sum = 0 i = 0 while i < n : sum = sum + a a = a + d i = i + 1 return sum # Driver function n = 20 a = 2.5 d = 1.5 print (sumOfAP(a, d, n)) # This code is contributed by Nikita Tiwari.
C#
// C# Program to find the sum of
// arithmetic series.
using System;
class GFG {
// Function to find sum of series.
static float sumOfAP(float a, float d,
int n)
{
float sum = 0;
for (int i = 0; i < n; i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
// Driver function
public static void Main()
{
int n = 20;
float a = 2.5f, d = 1.5f;
Console.Write(sumOfAP(a, d, n));
}
}
// This code is contributed by parashar.
PHP
<?php
// PHP Program to find the sum
// of arithmetic series.
// Function to find sum of series.
function sumOfAP($a, $d, $n)
{
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
$sum = $sum + $a;
$a = $a + $d;
}
return $sum;
}
// Driver Code
$n = 20;
$a = 2.5; $d = 1.5;
echo(sumOfAP($a, $d, $n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// Javascript Program to find the sum of arithmetic
// series.
// Function to find sum of series.
function sumOfAP(a, d, n)
{
let sum = 0;
for (let i=0;i<n;i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
// Driver function
let n = 20;
let a = 2.5, d = 1.5;
document.write(sumOfAP(a, d, n));
// This code is contributed by Mayank Tyagi
</script>
Producción:
335
Complejidad de tiempo: O (n)
Una solución eficiente para encontrar la suma de series aritméticas es usar la siguiente fórmula.
Sum of arithmetic series
= ((n / 2) * (2 * a + (n - 1) * d))
Where
a - First term
d - Common difference
n - No of terms
C++
// Efficient solution to find sum of arithmetic series.
#include<bits/stdc++.h>
using namespace std;
float sumOfAP(float a, float d, float n)
{
float sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}
// Driver code
int main()
{
float n = 20;
float a = 2.5, d = 1.5;
cout<<sumOfAP(a, d, n);
return 0;
}
Java
// Java Efficient solution to find
// sum of arithmetic series.
class GFG
{
static float sumOfAP(float a, float d, float n)
{
float sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}
// Driver code
public static void main (String[] args)
{
float n = 20;
float a = 2.5f, d = 1.5f;
System.out.print(sumOfAP(a, d, n));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 Efficient # solution to find sum # of arithmetic series. def sumOfAP(a, d, n): sum = (n / 2) * (2 * a + (n - 1) * d) return sum # Driver code n = 20 a = 2.5 d = 1.5 print(sumOfAP(a, d, n)) # This code is # contributed by sunnysingh
C#
// C# efficient solution to find
// sum of arithmetic series.
using System;
class GFG {
static float sumOfAP(float a,
float d,
float n)
{
float sum = (n / 2) *
(2 * a +
(n - 1) * d);
return sum;
}
// Driver code
static public void Main ()
{
float n = 20;
float a = 2.5f, d = 1.5f;
Console.WriteLine(sumOfAP(a, d, n));
}
}
// This code is contributed by Ajit.
PHP
<?php
// Efficient PHP code to find sum
// of arithmetic series.
// Function to find sum of series.
function sumOfAP($a, $d, $n)
{
$sum = ($n / 2) * (2 * $a +
($n - 1) * $d);
return $sum;
}
// Driver code
$n = 20;
$a = 2.5; $d = 1.5;
echo(sumOfAP($a, $d, $n));
// This code is contributed by Ajit.
?>
Javascript
// Efficient solution to find sum of arithmetic series.
function sumOfAP(a, d, n) {
let sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}
// Driver code
let n = 20;
let a = 2.5, d = 1.5;
document.write(sumOfAP(a, d, n));
// This code is contributed by Ashok
335
Complejidad de tiempo : O(1)
Complejidad del espacio : O(1) ya que usa solo variables constantes
¿Cómo funciona esta fórmula?
Podemos probar la fórmula usando inducción matemática. Podemos ver fácilmente que la fórmula se cumple para n = 1 y n = 2. Sea esto cierto para n = k-1.
Let the formula be true for n = k-1.
Sum of first k - 1 elements of geometric series is
= (((k-1))/ 2) * (2 * a + (k - 2) * d))
We know k-th term of arithmetic series is
= a + (k - 1)*d
Sum of first k elements =
= Sum of (k-1) numbers + k-th element
= (((k-1)/2)*(2*a + (k-2)*d)) + (a + (k-1)*d)
= [((k-1)(2a + (k-2)d) + (2a + 2kd - 2d)]/2
= ((k / 2) * (2 * a + (k - 1) * d))
Este artículo es una contribución de Dharmendra kumar . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA