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 <bits/stdc++.h>using namespace std;// function to calculate series sumint seriesSum(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 Codeint main(){ int n = 2; int a[] = { 1, 2, 3, 4 }; cout << seriesSum(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 seriesSum(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(seriesSum(n, a)); }} |
Python3
# Python3 program to find sum# of series with alternate signed # square AP sums.# Function to calculate series sumdef seriesSum(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 coden = 2a = [1, 2, 3, 4]print(seriesSum(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 seriesSum(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(seriesSum(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 sumfunction seriesSum($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; $a = array(1, 2, 3, 4); echo seriesSum($n, $a);// This code is contributed by anuj_67.?> |
Javascript
<script>// JavaScript program to find sum of // series with alternate signed // function to calculate // series sum function seriesSum(n, a) { let 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 let n = 2; let a = [1, 2, 3, 4]; document.write(seriesSum(n, a));// This code is contributed by code_hunt.</script> |
Output :
-10
Time complexity: O(2*n)
Auxiliary Space: O(1) since using constant space for variables
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 can 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 <bits/stdc++.h>using namespace std;// function to calculate // series sumint seriesSum(int n, int a[]){ return n * (a[0] * a[0] - a[2 * n - 1] * a[2 * n - 1]) / (2 * n - 1);}// Driver codeint main(){ int n = 2; int a[] = { 1, 2, 3, 4 }; cout << seriesSum(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 seriesSum(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(seriesSum(n, a)); }} |
Python3
# Efficient Python3 program # to find sum of series with # alternate signed square AP sums.# Function to calculate# series sumdef seriesSum(n, a): return (n * (a[0] * a[0] - a[2 * n - 1] * a[2 * n - 1]) / (2 * n - 1))# Driver coden = 2a = [1, 2, 3, 4] print(int(seriesSum(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 seriesSum(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(seriesSum(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 sumfunction seriesSum( $n, $a){ return $n * ($a[0] * $a[0] - $a[2 * $n - 1] * $a[2 * $n - 1]) / (2 * $n - 1);} // Driver code $n = 2; $a = array(1, 2, 3, 4); echo seriesSum($n, $a); // This code is contributed by anuj_67..?> |
Javascript
<script>// Efficient Javascript program to // find sum of series with // alternate signed square AP sums.// function to calculate// series sumfunction seriesSum(n, a){ return n * (a[0] * a[0] - a[2 * n - 1] * a[2 * n - 1]) / (2 * n - 1);} // Driver code let n = 2; a = [1, 2, 3, 4]; document.write(seriesSum(n, a)); // This code is contributed by _saurabh_jaiswal.</script> |
Output :
-10
Time complexity: O(1) since performing constant operations
Auxiliary Space: O(1) since using constant space for variables
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