Sum of series with alternate signed squares of AP

We are given the Integer n and also in the next line 2*n integers which represent a Arithmetic Progression series a₁, a₂, a₃…a_2n they are in AP. We need to find the sum of a₁² – a₂² + a₃²…. + a_2n-1² – a_2n² .

Examples :

Input : n = 2
        a[] = {1 2 3 4}
Output : -10
Explanation : 1² - 2² + 
3² 4² = -10.

Input : n = 3
        a[] = {2 4 6 8 10 12}
Output : -84

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach : We one by one find the sum of the square of the series with even terms negative and odd term as positive term .

C++

// CPP program to find sum of  
// series with alternate signed  
// square AP sums. 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to calculate series sum 
int seiresSum(int n, int a[]) 
{ 
    int res = 0; 
    for (int i = 0; i < 2 * n; i++)  
    { 
        if (i % 2 == 0) 
            res += a[i] * a[i]; 
        else
            res -= a[i] * a[i]; 
    } 
    return res; 
} 
  
// Driver Code 
int main() 
{ 
    int n = 2; 
    int a[] = { 1, 2, 3, 4 }; 
    cout << seiresSum(n, a); 
    return 0; 
} 

Java

// Java program to find sum of  
// series with alternate signed  
// square AP sums. 
import java.io.*; 
import java.lang.*; 
import java.util.*; 
  
class GFG  
{ 
  
    // function to calculate 
    // series sum 
    static int seiresSum(int n,  
                         int[] a) 
    { 
        int res = 0, i; 
        for (i = 0; i < 2 * n; i++)  
        { 
            if (i % 2 == 0) 
                res += a[i] * a[i]; 
            else
                res -= a[i] * a[i]; 
        } 
        return res; 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
        int n = 2; 
        int a[] = { 1, 2, 3, 4 }; 
        System.out.println(seiresSum(n, a)); 
    } 
} 

Python3

# Python3 program to find sum 
# of series with alternate signed   
# square AP sums. 
  
# Function to calculate series sum 
def seiresSum(n, a): 
    res = 0
      
    for i in range(0, 2 * n): 
        if (i % 2 == 0): 
            res += a[i] * a[i] 
        else: 
            res -= a[i] * a[i] 
    return res 
  
# Driver code 
n = 2
a = [1, 2, 3, 4] 
print(seiresSum(n, a)) 
  
# This code is contributed by Ajit. 

C#

// C# program to find sum of  
// series with alternate signed   
// square AP sums. 
using System; 
  
class GFG  
{ 
  
    // function to calculate  
    // series sum 
    static int seiresSum(int n,  
                         int[] a) 
    { 
        int res = 0, i; 
        for (i = 0; i < 2 * n; i++)  
        { 
            if (i % 2 == 0) 
                res += a[i] * a[i]; 
            else
                res -= a[i] * a[i]; 
        } 
        return res; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 2; 
        int []a = { 1, 2, 3, 4 }; 
        Console.WriteLine(seiresSum(n, a)); 
    } 
} 
  
//This code is contributed by vt_m. 

PHP

<?php 
// PHP program to find sum of  
// series with alternate signed   
// square AP sums. 
  
// function to calculate  
// series sum 
function seiresSum($n, $a) 
{ 
    $res = 0; 
    for ( $i = 0; $i < 2 * $n; $i++) 
    { 
        if ($i % 2 == 0) 
            $res += $a[$i] * $a[$i]; 
        else
            $res -= $a[$i] * $a[$i]; 
    } 
    return $res; 
} 
  
    // Driver Code 
    $n = 2; 
    $a = array(1, 2, 3, 4); 
    echo seiresSum($n, $a); 
  
// This code is contributed by anuj_67. 
?> 

Output :

-10

Efficient Approach:Use of Arithmetic progression Application

We know that common difference d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃

Result = a₁² – a₂² + a₃²…. + a_2n-1² – a_2n²

= (a₁ – a₂)*(a₁ + a₂) + (a₃ – a₄)*(a₃ +a₄)+….+(a_2n-1 – a_2n)*(a_2n-1 + a_2n)

So as common difference is common to the series then :
(a₁ – a₂)[a₁ + a₂ + a₃…a_2n]

now we can write :

(-d)*(Sum of the term of the 2n term of AP)
(-d)*[((2*n)*(a₁ + a_2n))/2]
now we know that common difference is : d = (a₁ - a₂)
Then the difference between : g = (a_2n - a₁)
So we can conclude that g = d*(2*n - 1)
the we ca replace d by : g/(2*n - 1)

So our result becomes : (n/(2*n - 1)) * (a₁² - a_2n²)

C++

// Efficient CPP program to  
// find sum of series with  
// alternate signed square AP sums. 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to calculate  
// series sum 
int seiresSum(int n, int a[]) 
{ 
    return n * (a[0] * a[0] - a[2 * n - 1] *  
                a[2 * n - 1]) / (2 * n - 1); 
} 
  
// Driver code 
int main() 
{ 
    int n = 2; 
    int a[] = { 1, 2, 3, 4 }; 
    cout << seiresSum(n, a); 
    return 0; 
} 

Java

// Efficient Java program to   
// find sum of series with  
// alternate signed square AP sums. 
import java.io.*; 
import java.lang.*; 
import java.util.*; 
  
class GFG  
{ 
    static int seiresSum(int n,  
                         int[] a) 
    { 
    return n * (a[0] * a[0] - a[2 * n - 1] * 
                a[2 * n - 1]) / (2 * n - 1); 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        int n = 2; 
        int a[] = { 1, 2, 3, 4 }; 
        System.out.println(seiresSum(n, a)); 
    } 
} 

Python3

   
# Efficient Python3 program   
# to find sum of series with   
# alternate signed square AP sums. 
  
# Function to calculate 
# series sum 
def seiresSum(n, a): 
  
    return (n * (a[0] * a[0] - a[2 * n - 1] * 
                 a[2 * n - 1]) / (2 * n - 1)) 
  
# Driver code 
n = 2
a = [1, 2, 3, 4]  
print(int(seiresSum(n, a))) 
  
# This code is contributed  
# by Smitha Dinesh Semwal. 

C#

// Efficient C# program to find sum  
// of series with alternate signed  
// square AP sums. 
using System; 
  
class GFG  
{ 
    static int seiresSum(int n, int[] a) 
    { 
    return n * (a[0] * a[0] - a[2 * n - 1] * 
                a[2 * n - 1]) / (2 * n - 1); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        int n = 2; 
        int []a= { 1, 2, 3, 4 }; 
        Console.WriteLine(seiresSum(n, a)); 
    } 
} 
  
// This code is contributed by anuj_67.. 

PHP

<?php 
// Efficient PHP program to  
// find sum of series with  
// alternate signed square AP sums. 
  
// function to calculate 
// series sum 
function seiresSum( $n, $a) 
{ 
    return $n * ($a[0] * $a[0] -  
                 $a[2 * $n - 1] * 
                 $a[2 * $n - 1]) /  
                 (2 * $n - 1); 
} 
  
    // Driver code 
    $n = 2; 
    $a = array(1, 2, 3, 4); 
    echo seiresSum($n, $a); 
      
// This code is contributed by anuj_67.. 
?> 

Output :

-10

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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