# Sum of series with alternate signed squares of AP

• Difficulty Level : Hard
• Last Updated : 19 Apr, 2021

We are given the Integer n and also in the next line 2*n integers which represent a Arithmetic Progression series a1, a2, a3…a2n they are in AP. We need to find the sum of a12 – a22 + a32…. + a2n-12 – a2n2 .
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```

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 ``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

 ``

## Javascript

 ``
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…. + a2n-12 – a2n2
= (a1 – a2)*(a1 + a2) + (a3 – a4)*(a3 +a4)+….+(a2n-1 – a2n)*(a2n-1 + a2n)
So as common difference is common to the series then :
(a1 – a2)[a1 + a2 + a3…a2n]
now we can write :

```(-d)*(Sum of the term of the 2n term of AP)
(-d)*[((2*n)*(a1 + a2n))/2]
now we know that common difference is : d = (a1 - a2)
Then the difference between : g = (a2n - 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 - a2n2)```

## C++

 `// Efficient CPP program to``// find sum of series with``// alternate signed square AP sums.``#include ``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

 ``

## Javascript

 ``
Output :
`-10`

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