# Sum of elements of an AP in the given range

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

1. Multiply the first element of the range to the number of elements in the range.
2. 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 be 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 ` `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 - arr; ` ` `  `    ``// 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, 4 }, ` `                          ``{ 2, 6 }, ` `                          ``{ 5, 6 } }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``for` `(``int` `i = 0; i < queries; i++) ` `        ``cout << findSum(arr, n, q[i], q[i]) ` `             ``<< 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#

 `// 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 - arr; ` `   `  `    ``// 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 `

Output:

```18
40
22
```

Time complexity: O(1)
Space complexity: O(1)

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