Sum of elements of an AP in the given range

Last Updated : 12 Jul, 2025

Given an arithmetic series in arr and Q queries in the form of [L, R], where L is the left boundary of the range and R is the right boundary. The task is to find the sum of the AP elements in the given range.
Note: The range is 1-indexed and 1 ? L, R ? N, where N is the size of arr.

Examples:

Input: arr[] = {2, 4, 6, 8, 10, 12, 14, 16}, Q = [[2, 4], [2, 6], [5, 8]]
Output:
18
40
52
Explanation:
Range 1: arr = {4, 6, 8}. Therefore sum = 18
Range 2: arr = {4, 6, 8, 10, 12}. Therefore sum = 40
Range 3: arr = {10, 12, 14, 16}. Therefore sum = 52

Input: arr[] = {7, 14, 21, 28, 35, 42}, Q = [[1, 6], [2, 4], [3, 3]]
Output:
147
63
21
Explanation:
Range 1: arr = {7, 14, 21, 28, 35, 42}. Therefore sum = 147
Range 2: arr = {14, 21, 28}. Therefore sum = 63
Range 3: arr = {21}. Therefore sum = 21

Approach: Since the given sequence is an arithmetic progression, the sum can be easily found in two steps efficiently:

Multiply the first element of the range by the number of elements in the range.
Add (d*k*(k+1))/2 to it, where d is the common difference of the AP and k is (number of elements in the range - 1) which corresponds to the number of gaps.

For example:
Suppose a[i] be the first element of the range, d is the common difference of AP and k+1 be the number of elements in the given range.
Then the sum of the range would be

= a[i] + a[i+1] + a[i+2] + ..... + a[i+k]
= a[i] + a[i] + d + a[i] + 2 * d + .... + a[i] + k * d
= a[i] * (k + 1) + d * (1 + 2 + ... + k)
= a[i] * (k + 1) + (d * k * (k+1))/2

Below is the implementation of the above approach:

CPP

// C++ program to find the sum of elements
// of an AP in the given range

#include <bits/stdc++.h>
using namespace std;

// Function to find sum in the given range
int findSum(int* arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;

    // Find the common difference
    int d = arr[1] - arr[0];

    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;

    return ans;
}

// Driver code
int main()
{
    int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int q[queries][2] = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = sizeof(arr) / sizeof(arr[0]);

    for (int i = 0; i < queries; i++)
        cout << findSum(arr, n, q[i][0], q[i][1])
             << endl;
}

Java

// Java program to find the sum of elements
// of an AP in the given range
class GFG{
 
// Function to find sum in the given range
static int findSum(int []arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;
 
    // Find the common difference
    int d = arr[1] - arr[0];
 
    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
 
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int q[][] = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = arr.length;
 
    for (int i = 0; i < queries; i++)
        System.out.print(findSum(arr, n, q[i][0], q[i][1])
             +"\n");
}
}

// This code is contributed by Princi Singh

Python3

# Python program to find the sum of elements
# of an AP in the given range

# Function to find sum in the given range
def findSum(arr, n, left, right):

    # Find the value of k
    k = right - left;

    # Find the common difference
    d = arr[1] - arr[0];

    # Find the sum
    ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) // 2;

    return ans;

# Driver code
if __name__ == '__main__':
    arr = [ 2, 4, 6, 8, 10, 12, 14, 16 ];
    queries = 3;
    q = [[ 2, 4 ],[ 2, 6 ],[ 5, 6 ]];
    n = len(arr);

    for i in range(queries):
        print(findSum(arr, n, q[i][0], q[i][1]));

# This code is contributed by sapnasingh4991

// C# program to find the sum of elements
// of an AP in the given range
using System;

class GFG{
  
// Function to find sum in the given range
static int findSum(int []arr, int n,
            int left, int right)
{
    // Find the value of k
    int k = right - left;
  
    // Find the common difference
    int d = arr[1] - arr[0];
  
    // Find the sum
    int ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
  
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
    int queries = 3;
    int [,]q = { { 2, 4 },
                          { 2, 6 },
                          { 5, 6 } };
    int n = arr.Length;
  
    for (int i = 0; i < queries; i++)
        Console.Write(findSum(arr, n, q[i,0], q[i,1])
             +"\n");
}
}

// This code is contributed by PrinciRaj1992

JavaScript

<script>

// JavaScript program to find the sum of elements
// of an AP in the given range

// Function to find sum in the given range
function findSum(arr, n, left, right)
{
    // Find the value of k
    let k = right - left;
   
    // Find the common difference
    let d = arr[1] - arr[0];
   
    // Find the sum
    let ans = arr[left - 1] * (k + 1);
    ans = ans + (d * (k * (k + 1))) / 2;
   
    return ans;
}

// Driver Code

    let arr = [ 2, 4, 6, 8, 10, 12, 14, 16 ];
    let queries = 3;
    let q = [[ 2, 4 ],
             [ 2, 6 ],
             [ 5, 6 ]];
    let n = arr.length;
   
    for (let i = 0; i < queries; i++)
        document.write(findSum(arr, n, q[i][0], q[i][1])
             +"<br/>");
    
</script>

Output

18
40
22

Time complexity: O(q), where q is the length of the given queries array.
Auxiliary Space: O(1)

Method 2: Using Prefix Sum

Approach:

Create a prefix sum array for the given arithmetic series.
For each query, calculate the sum of the range by subtracting the prefix sum at L-1 from the prefix sum at R.
Output the sum for each query.

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;

vector<int> arithmetic_series_sum(int arr[], int N, vector<vector<int>> queries) {
    // precompute prefix sum
    int prefixSum[N];
    prefixSum[0] = arr[0];
    for (int i = 1; i < N; i++) {
        prefixSum[i] = prefixSum[i - 1] + arr[i];
    }

    // process queries
    vector<int> ans;
    for (int i = 0; i < queries.size(); i++) {
        int L = queries[i][0];
        int R = queries[i][1];
         int sum = prefixSum[R - 1] - prefixSum[L - 2];
        ans.push_back(sum);
    }

    return ans;
}

int main() {
    int arr[] = {2, 4, 6, 8, 10, 12, 14, 16};
    int N = sizeof(arr) / sizeof(arr[0]);

    // define queries
    vector<vector<int>> queries = {{2, 4},{2, 6},{5, 6}};

    // call function to compute arithmetic series sum for queries
    vector<int> ans = arithmetic_series_sum(arr, N, queries);

    // print answers
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << endl;
    }

    return 0;
}

Java

import java.util.ArrayList;

public class GFG {
    public static ArrayList<Integer> arithmeticSeriesSum(
        int[] arr, int N,
        ArrayList<ArrayList<Integer> > queries)
    {
        // Precompute prefix sum
        int[] prefixSum = new int[N];
        prefixSum[0] = arr[0];
        for (int i = 1; i < N; i++) {
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        }

        // Process queries
        ArrayList<Integer> ans = new ArrayList<>();
        for (int i = 0; i < queries.size(); i++) {
            int L = queries.get(i).get(0);
            int R = queries.get(i).get(1);
            int sum = prefixSum[R - 1]
                      - (L > 1 ? prefixSum[L - 2] : 0);
            ans.add(sum);
        }

        return ans;
    }

    public static void main(String[] args)
    {
        int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
        int N = arr.length;

        // Define queries
        ArrayList<ArrayList<Integer> > queries
            = new ArrayList<>();
        queries.add(
            new ArrayList<>(java.util.Arrays.asList(2, 4)));
        queries.add(
            new ArrayList<>(java.util.Arrays.asList(2, 6)));
        queries.add(
            new ArrayList<>(java.util.Arrays.asList(5, 6)));

        // Call function to compute arithmetic series sum
        // for queries
        ArrayList<Integer> ans
            = arithmeticSeriesSum(arr, N, queries);

        // Print answers
        for (int i = 0; i < ans.size(); i++) {
            System.out.println(ans.get(i));
        }
    }
}

Python3

def arithmetic_series_sum(arr, N, queries):
    # precompute prefix sum
    prefixSum = [0] * N
    prefixSum[0] = arr[0]
    for i in range(1, N):
        prefixSum[i] = prefixSum[i - 1] + arr[i]

    # process queries
    ans = []
    for i in range(len(queries)):
        L = queries[i][0]
        R = queries[i][1]
        # Compute the sum using prefix sums
        sum = prefixSum[R - 1] - prefixSum[L - 2]
        ans.append(sum)

    return ans


if __name__ == "__main__":
    arr = [2, 4, 6, 8, 10, 12, 14, 16]
    N = len(arr)

    # define queries
    queries = [[2, 4], [2, 6], [5, 6]]

    # call function to compute arithmetic series sum for queries
    ans = arithmetic_series_sum(arr, N, queries)

    # print answers
    for i in range(len(ans)):
        print(ans[i])

using System;
using System.Collections.Generic;

class GFG
{
    static List<int> ArithmeticSeriesSum(int[] arr, int N, List<List<int>> queries)
    {
        // precompute prefix sum
        int[] prefixSum = new int[N];
        prefixSum[0] = arr[0];
        for (int i = 1; i < N; i++)
        {
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        }

        // process queries
        List<int> ans = new List<int>();
        for (int i = 0; i < queries.Count; i++)
        {
            int L = queries[i][0];
            int R = queries[i][1];
            int sum = prefixSum[R - 1] - (L - 2 >= 0 ? prefixSum[L - 2] : 0);
            ans.Add(sum);
        }

        return ans;
    }

    static void Main()
    {
        int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
        int N = arr.Length;

        // define queries
        List<List<int>> queries = new List<List<int>> {
            new List<int> { 2, 4 },
            new List<int> { 2, 6 },
            new List<int> { 5, 6 }
        };

        // call function to compute arithmetic series sum for queries
        List<int> ans = ArithmeticSeriesSum(arr, N, queries);

        // print answers
        foreach (int value in ans)
        {
            Console.WriteLine(value);
        }
    }
}

JavaScript

function arithmeticSeriesSum(arr, N, queries) {
    // precompute prefix sum
    let prefixSum = [];
    prefixSum[0] = arr[0];
    for (let i = 1; i < N; i++) {
        prefixSum[i] = prefixSum[i - 1] + arr[i];
    }

    // process queries
    let ans = [];
    for (let i = 0; i < queries.length; i++) {
        let L = queries[i][0];
        let R = queries[i][1];
        let sum = prefixSum[R - 1] - (L > 1 ? prefixSum[L - 2] : 0);
        ans.push(sum);
    }

    return ans;
}

let arr = [2, 4, 6, 8, 10, 12, 14, 16];
let N = arr.length;

// define queries
let queries = [[2, 4], [2, 6], [5, 6]];

// call function to compute arithmetic series sum for queries
let ans = arithmeticSeriesSum(arr, N, queries);

// print answers
for (let i = 0; i < ans.length; i++) {
    console.log(ans[i]);
}

Output

18
40
22

Time complexity: O(N + Q),N is the size of the arithmetic series and Q is the number of queries.
Auxiliary Space: O(N) for the prefix sum array.

Analysis of Algorithms

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Sum of elements of an AP in the given range

Method 2: Using Prefix Sum

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