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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int n = 2;
int a[] = { 1, 2, 3, 4 };
cout << seiresSum(n, a);
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
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;
}
public static void main(String args[])
{
int n = 2;
int a[] = { 1, 2, 3, 4 };
System.out.println(seiresSum(n, a));
}
}
|
Python3
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
n = 2
a = [1, 2, 3, 4]
print(seiresSum(n, a))
|
C#
using System;
class GFG
{
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;
}
public static void Main()
{
int n = 2;
int []a = { 1, 2, 3, 4 };
Console.WriteLine(seiresSum(n, a));
}
}
|
PHP
<?php
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;
}
$n = 2;
$a = array(1, 2, 3, 4);
echo seiresSum($n, $a);
?>
|
Javascript
<script>
function seiresSum(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;
}
let n = 2;
let a = [1, 2, 3, 4];
document.write(seiresSum(n, a));
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
int seiresSum(int n, int a[])
{
return n * (a[0] * a[0] - a[2 * n - 1] *
a[2 * n - 1]) / (2 * n - 1);
}
int main()
{
int n = 2;
int a[] = { 1, 2, 3, 4 };
cout << seiresSum(n, a);
return 0;
}
|
Java
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);
}
public static void main(String args[])
{
int n = 2;
int a[] = { 1, 2, 3, 4 };
System.out.println(seiresSum(n, a));
}
}
|
Python3
def seiresSum(n, a):
return (n * (a[0] * a[0] - a[2 * n - 1] *
a[2 * n - 1]) / (2 * n - 1))
n = 2
a = [1, 2, 3, 4]
print(int(seiresSum(n, a)))
|
C#
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);
}
public static void Main()
{
int n = 2;
int []a= { 1, 2, 3, 4 };
Console.WriteLine(seiresSum(n, a));
}
}
|
PHP
<?php
function seiresSum( $n, $a)
{
return $n * ($a[0] * $a[0] -
$a[2 * $n - 1] *
$a[2 * $n - 1]) /
(2 * $n - 1);
}
$n = 2;
$a = array(1, 2, 3, 4);
echo seiresSum($n, $a);
?>
|
Javascript
<script>
function seiresSum(n, a)
{
return n * (a[0] * a[0] -
a[2 * n - 1] *
a[2 * n - 1]) /
(2 * n - 1);
}
let n = 2;
a = [1, 2, 3, 4];
document.write(seiresSum(n, a));
</script>
|
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