Sum of area of alternative concentric circle with radius 1,2,3,4……….N

Given that Concentric circle of radii 1,2,3,4……….N cm are drawn. The interior of the smallest circle is colored white and angular regions are colored alternately green and white so that no two adjacent regions are of the same color. The task is to find the total area of the green region in sq cm.

Examples:

Input: N=100
Output: 15865.042900628456

Input: N=10
Output: 172.78759594743863

Approach: Area of the first green region would be-

[π(R2²- R1²)]

Similarly, the area of all alternative green circles would be-

[π(R4²- R3²)], [π(R6²- R5²)]…..

The total area of the green region is the sum of the area of the green region is-

A  =  [ π { (R2² – R1²) + (R4² – R3²) + (R6² – R5²) …………. (R(N)² –  R(N-1)²}]

A =   [ π { ((R2 – R1)(R2 + R1)) + ((R4 – R3)(R4 + R3)) + ((R6 – R5)(R6 + R5)) …………. ((RN –  R(N – 1)(RN + R(N – 1)) } ]

Since the difference between the radius of 2 concentric circles is only 1 cm so R(N) – R(N – 1) = 1. Hence,

A  =   [ π {R1 + R2 + R3 + R4 + R5 + R6 ………….R(N-1) + R(N) } ]

(R1 + R2 + R3 + R4 + R5 + R6 ………….R(N-1) + R(N) forms an arithmetic progression so we have to find the sum of arithmetic progression with starting radius as 1 and last radius as N with a common difference of 1.

Sum of A.P. is-

S_N = N * (2 * a +(N – 1) * d) / 2  

or 

S_N = N * (a + l) / 2

Thus, the area of the green region is-

A = Sn * π 

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate area
// of concentric circles
double area(int N, int a, int l)
{

    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;

    // return area
    return S * M_PI;
}

// Driver code
int main()
{

    // N is number of concentric circles
    int N = 100;

    // Radius of interior circle
    int a = 1;

    // l radius of exterior circle
    int l = N;

    // r is result
    double r = area(N, a, l);

    // Print result
    cout << fixed << setprecision(12) << r;
}

// This code is contributed by Samim Hossain Mondal.

Java

// Java code for the above approach
class GFG
{

// Function to calculate area
// of concentric circles
static double area(int N, int a, int l)
{

    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;

    // return area
    return S * Math.PI;
}

// Driver code
public static void main(String args[])
{

    // N is number of concentric circles
    int N = 100;

    // Radius of interior circle
    int a = 1;

    // l radius of exterior circle
    int l = N;

    // r is result
    double r = area(N, a, l);

    // Print result
    System.out.println(r);
}
}

// This code is contributed by gfgking

Python3

# import pi from math module
from math import pi

# Function to calculate area
# of concentric circles
def area(N,a):

    # Formula to find sum of
    # Arithmetic Progression
    S = N * (a + l) / 2

    # return area
    return S * pi

# N is number of concentric circles
N = 100

# Radius of interior circle
a = 1

# l radius of exterior circle
l = N

# r is result
r = area(N, a)

# Print result
print(r)

C#

// C# program for the above approach
using System;
class GFG
{

  // Function to calculate area
  // of concentric circles
  static double area(int N, int a, int l)
  {

    // Formula to find sum of
    // Arithmetic Progression
    double S = N * (a + l) / 2;

    // return area
    return S * Math.PI;
  }

  // Driver Code:
  public static void Main()
  {
    // N is number of concentric circles
    int N = 100;

    // Radius of interior circle
    int a = 1;

    // l radius of exterior circle
    int l = N;

    // r is result
    double r = area(N, a, l);

    // Print result
    Console.WriteLine(r);
  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
       // JavaScript code for the above approach

       // Function to calculate area
       // of concentric circles
       function area(N, a) {

           // Formula to find sum of
           // Arithmetic Progression
           S = N * (a + l) / 2

           // return area
           return S * Math.PI
       }

       // N is number of concentric circles
       N = 100

       // Radius of interior circle
       a = 1

       // l radius of exterior circle
       l = N

       // r is result
       r = area(N, a)

       // Print result
       document.write(r)

 // This code is contributed by Potta Lokesh
   </script>

Output

15865.042900628456

Time Complexity: O(1) 
Auxiliary Space: O(1)