# Find the missing number in Arithmetic Progression

• Difficulty Level : Medium
• Last Updated : 03 Nov, 2021

Given an array that represents elements of arithmetic progression in order. One element is missing in the progression, find the missing number.

Examples:

```Input: arr[]  = {2, 4, 8, 10, 12, 14}
Output: 6

Input: arr[]  = {1, 6, 11, 16, 21, 31};
Output: 26```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A Simple Solution is to linearly traverse the array and find the missing number. Time complexity of this solution is O(n). We can solve this problem in O(Logn) time using Binary Search. The idea is to go to the middle element. Check if the difference between middle and next to middle is equal to diff or not, if not then the missing element lies between mid and mid+1. If the middle element is equal to n/2th term in Arithmetic Series (Let n be the number of elements in input array), then missing element lies in right half. Else element lies in left half.

Following is implementation of above idea.

## C++

```// C++ program to find the missing number
// in a given arithmetic progression
#include<iostream>
using namespace std;
#define INT_MAX 2147483647;
class GFG
{

// A binary search based recursive function that returns
// the missing element in arithmetic progression
public:int findMissingUtil(int arr[], int low,
int high, int diff)
{
// There must be two elements to find the missing
if (high <= low)
return INT_MAX;

// Find index of middle element
int mid = low + (high - low) / 2;

// The element just after the middle element is missing.
// The arr[mid+1] must exist, because we return when
// (low == high) and take floor of (high-low)/2
if (arr[mid + 1] - arr[mid] != diff)
return (arr[mid] + diff);

// The element just before mid is missing
if (mid > 0 && arr[mid] - arr[mid - 1] != diff)
return (arr[mid - 1] + diff);

// If the elements till mid follow AP, then recur
// for right half
if (arr[mid] == arr[0] + mid * diff)
return findMissingUtil(arr, mid + 1,
high, diff);

// Else recur for left half
return findMissingUtil(arr, low, mid - 1, diff);
}

// The function uses findMissingUtil() to
// find the missing element in AP.
// It assumes that there is exactly one
// missing element and may give incorrect result
// when there is no missing element or
// more than one missing elements. This function
// also assumes that the difference in AP is an
// integer.
int findMissing(int arr[], int n)
{
// If exactly one element is missing, then we can find
// difference of arithmetic progression using following
// formula. Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the difference is
// an integer.
int diff = (arr[n - 1] - arr[0]) / n;

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}
};

// Driver Code
int main()
{
GFG g;
int arr[] = {2, 4, 8, 10, 12, 14};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "The missing element is "
<< g.findMissing(arr, n);
return 0;
}

// This code is contributed by Soumik
```

## C

```// A C program to find the missing number in a given
// arithmetic progression
#include <stdio.h>
#include <limits.h>

// A binary search based recursive function that returns
// the missing element in arithmetic progression
int findMissingUtil(int arr[], int low, int high, int diff)
{
// There must be two elements to find the missing
if (high <= low)
return INT_MAX;

// Find index of middle element
int mid = low + (high - low)/2;

// The element just after the middle element is missing.
// The arr[mid+1] must exist, because we return when
// (low == high) and take floor of (high-low)/2
if (arr[mid+1] - arr[mid] != diff)
return (arr[mid] + diff);

// The element just before mid is missing
if (mid > 0 && arr[mid] - arr[mid-1] != diff)
return (arr[mid-1] + diff);

// If the elements till mid follow AP, then recur
// for right half
if (arr[mid] == arr[0] + mid*diff)
return findMissingUtil(arr, mid+1, high, diff);

// Else recur for left half
return findMissingUtil(arr, low, mid-1, diff);
}

// The function uses findMissingUtil() to find the missing
// element in AP. It assumes that there is exactly one missing
// element and may give incorrect result when there is no missing
// element or more than one missing elements.
// This function also assumes that the difference in AP is an
// integer.
int findMissing(int arr[], int n)
{
// If exactly one element is missing, then we can find
// difference of arithmetic progression using following
// formula. Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the difference is
// an integer.
int diff = (arr[n-1] - arr[0])/n;

// Binary search for the missing number using above
// calculated diff
return findMissingUtil(arr, 0, n-1, diff);
}

/* Driver program to check above functions */
int main()
{
int arr[] = {2, 4, 8, 10, 12, 14};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The missing element is %d", findMissing(arr, n));
return 0;
}
```

## Java

```// A Java program to find
// the missing number in
// a given arithmetic
// progression
import java.io.*;

class GFG
{

// A binary search based
// recursive function that
// returns the missing
// element in arithmetic
// progression
static int findMissingUtil(int arr[], int low,
int high, int diff)
{
// There must be two elements
// to find the missing
if (high <= low)
return Integer.MAX_VALUE;

// Find index of
// middle element
int mid = low + (high - low) / 2;

// The element just after the
// middle element is missing.
// The arr[mid+1] must exist,
// because we return when
// (low == high) and take
// floor of (high-low)/2
if (arr[mid + 1] - arr[mid] != diff)
return (arr[mid] + diff);

// The element just
// before mid is missing
if (mid > 0 && arr[mid] -
arr[mid - 1] != diff)
return (arr[mid - 1] + diff);

// If the elements till mid follow
// AP, then recur for right half
if (arr[mid] == arr[0] + mid * diff)
return findMissingUtil(arr, mid + 1,
high, diff);

// Else recur for left half
return findMissingUtil(arr, low, mid - 1, diff);
}

// The function uses findMissingUtil()
// to find the missing element in AP.
// It assumes that there is exactly
// one missing element and may give
// incorrect result when there is no
// missing element or more than one
// missing elements. This function also
// assumes that the difference in AP is
// an integer.
static int findMissing(int arr[], int n)
{
// If exactly one element is missing,
// then we can find difference of
// arithmetic progression using
// following formula. Example, 2, 4,
// 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that
// the difference is an integer.
int diff = (arr[n - 1] - arr[0]) / n;

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}

// Driver Code
public static void main (String[] args)
{
int arr[] = {2, 4, 8, 10, 12, 14};
int n = arr.length;
System.out.println("The missing element is "+
findMissing(arr, n));
}
}

// This code is contributed by anuj_67.
```

## Python3

```# A Python3 program to find the missing
# number in a given arithmetic progression
import sys

# A binary search based recursive function
# that returns the missing element in
# arithmetic progression
def findMissingUtil(arr, low, high, diff):

# There must be two elements to
# find the missing
if (high <= low):
return sys.maxsize;

# Find index of middle element
mid = int(low + (high - low) / 2);

# The element just after the middle
# element is missing. The arr[mid+1]
# must exist, because we return when
# (low == high) and take floor of
# (high-low)/2
if (arr[mid + 1] - arr[mid] != diff):
return (arr[mid] + diff);

# The element just before mid is missing
if (mid > 0 and arr[mid] -
arr[mid - 1] != diff):
return (arr[mid - 1] + diff);

# If the elements till mid follow AP,
# then recur for right half
if (arr[mid] == arr[0] + mid * diff):
return findMissingUtil(arr, mid + 1,
high, diff);

# Else recur for left half
return findMissingUtil(arr, low,
mid - 1, diff);

# The function uses findMissingUtil() to find
# the missing element in AP. It assumes that
# there is exactly one missing element and may
# give incorrect result when there is no missing
# element or more than one missing elements.
# This function also assumes that the difference
# in AP is an integer.
def findMissing(arr, n):

# If exactly one element is missing, then
# we can find difference of arithmetic
# progression using following formula.
# Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
# The assumption in formula is that the
# difference is an integer.
diff = int((arr[n - 1] - arr[0]) / n);

# Binary search for the missing number
# using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);

# Driver Code
arr = [2, 4, 8, 10, 12, 14];
n = len(arr);
print("The missing element is",
findMissing(arr, n));

# This code is contributed by chandan_jnu
```

## C#

```// A C# program to find
// the missing number in
// a given arithmetic
// progression
using System;

class GFG
{

// A binary search based
// recursive function that
// returns the missing
// element in arithmetic
// progression
static int findMissingUtil(int []arr, int low,
int high, int diff)
{
// There must be two elements
// to find the missing
if (high <= low)
return int.MaxValue;

// Find index of
// middle element
int mid = low + (high -
low) / 2;

// The element just after the
// middle element is missing.
// The arr[mid+1] must exist,
// because we return when
// (low == high) and take
// floor of (high-low)/2
if (arr[mid + 1] -
arr[mid] != diff)
return (arr[mid] + diff);

// The element just
// before mid is missing
if (mid > 0 && arr[mid] -
arr[mid - 1] != diff)
return (arr[mid - 1] + diff);

// If the elements till mid follow
// AP, then recur for right half
if (arr[mid] == arr[0] +
mid * diff)
return findMissingUtil(arr, mid + 1,
high, diff);

// Else recur for left half
return findMissingUtil(arr, low,
mid - 1, diff);
}

// The function uses findMissingUtil()
// to find the missing element
// in AP. It assumes that there
// is exactly one missing element
// and may give incorrect result
// when there is no missing element
// or more than one missing elements.
// This function also assumes that
// the difference in AP is an integer.
static int findMissing(int []arr, int n)
{
// If exactly one element
// is missing, then we can
// find difference of arithmetic
// progression using following
// formula. Example, 2, 4, 6, 10,
// diff = (10-2)/4 = 2.The assumption
// in formula is that the difference
// is an integer.
int diff = (arr[n - 1] -
arr[0]) / n;

// Binary search for the
// missing number using
// above calculated diff
return findMissingUtil(arr, 0,
n - 1, diff);
}

// Driver Code
public static void Main ()
{
int []arr = {2, 4, 8,
10, 12, 14};
int n = arr.Length;
Console.WriteLine("The missing element is "+
findMissing(arr, n));
}
}

// This code is contributed by anuj_67.
```

## PHP

```<?php
// A PHP program to find the missing
// number in a given arithmetic progression

// A binary search based recursive function
// that returns the missing element in
// arithmetic progression

function findMissingUtil(\$arr, \$low, \$high, \$diff)
{
// There must be two elements to
// find the missing
if (\$high <= \$low)
return PHP_INT_MAX;

// Find index of middle element
\$mid = \$low + (\$high - \$low) / 2;

// The element just after the middle
// element is missing. The arr[mid+1]
// must exist, because we return when
// (low == high) and take floor of (high-low)/2
if (\$arr[\$mid + 1] - \$arr[\$mid] != \$diff)
return (\$arr[\$mid] + \$diff);

// The element just before mid is missing
if (\$mid > 0 && \$arr[\$mid] - \$arr[\$mid - 1] != \$diff)
return (\$arr[\$mid - 1] + \$diff);

// If the elements till mid follow AP,
// then recur for right half
if (\$arr[\$mid] == \$arr[0] + \$mid * \$diff)
return findMissingUtil(\$arr, \$mid + 1,
\$high, \$diff);

// Else recur for left half
return findMissingUtil(\$arr, \$low, \$mid - 1, \$diff);
}

// The function uses findMissingUtil() to find
// the missing element in AP. It assumes that
// there is exactly one missing element and may
// give incorrect result when there is no missing
// element or more than one missing elements.
// This function also assumes that the difference
// in AP is an integer.
function findMissing(\$arr, \$n)
{
// If exactly one element is missing, then
// we can find difference of arithmetic
// progression using following formula.
// Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the
// difference is an integer.
\$diff = (\$arr[\$n - 1] - \$arr[0]) / \$n;

// Binary search for the missing number
// using above calculated diff
return findMissingUtil(\$arr, 0, \$n - 1, \$diff);
}

// Driver Code
\$arr = array(2, 4, 8, 10, 12, 14);
\$n = sizeof(\$arr);
echo "The missing element is ",
findMissing(\$arr, \$n);

// This code is contributed by Sach_Code
?>
```

## Javascript

```<script>

// A JavaScript program to find
// the missing number in
// a given arithmetic
// progression

// A binary search based
// recursive function that
// returns the missing
// element in arithmetic
// progression
function findMissingUtil(arr, low, high, diff)
{
// There must be two elements
// to find the missing
if (high <= low)
return Number.MAX_VALUE;

// Find index of
// middle element
let mid = low + parseInt((high - low) / 2, 10);

// The element just after the
// middle element is missing.
// The arr[mid+1] must exist,
// because we return when
// (low == high) and take
// floor of (high-low)/2
if (arr[mid + 1] - arr[mid] != diff)
return (arr[mid] + diff);

// The element just
// before mid is missing
if (mid > 0 && arr[mid] -
arr[mid - 1] != diff)
return (arr[mid - 1] + diff);

// If the elements till mid follow
// AP, then recur for right half
if (arr[mid] == arr[0] + mid * diff)
return findMissingUtil(arr, mid + 1, high, diff);

// Else recur for left half
return findMissingUtil(arr, low, mid - 1, diff);
}

// The function uses findMissingUtil()
// to find the missing element in AP.
// It assumes that there is exactly
// one missing element and may give
// incorrect result when there is no
// missing element or more than one
// missing elements. This function also
// assumes that the difference in AP is
// an integer.
function findMissing(arr, n)
{
// If exactly one element is missing,
// then we can find difference of
// arithmetic progression using
// following formula. Example, 2, 4,
// 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that
// the difference is an integer.
let diff = parseInt((arr[n - 1] - arr[0]) / n, 10);

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}

let arr = [2, 4, 8, 10, 12, 14];
let n = arr.length;
document.write("The missing element is "+
findMissing(arr, n));

</script>
```
Output
`The missing element is 6`

Time Complexity: O(log n)

Auxiliary Space: O(1)

Iterative:

The idea is to go to the middle element. Check if the index of middle element is equal to (nth position of middle element in AP) – 1 then the missing element lies at right half if not then the missing element lies at left half (this idea is similar to Find the only repeating element in a sorted array of size n ).  After breaking out of binary search loop the missing element will lie between high and low. We can find the missing element by adding a common difference with element at index high or by subtracting a common difference with element at index low.

Following is implementation of above idea.

## C++

```// C++ program to find the missing number
// in a given arithmetic progression
#include<iostream>
using namespace std;
#define INT_MAX 2147483647;
class GFG
{

// A binary search based function that returns
// the missing element in arithmetic progression
public:int findMissingUtil(int arr[], int low,
int high, int diff)
{
// Find index of middle element
int mid;
while (low <= high)
{
// find index of middle element
mid = (low + high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((arr[mid] - arr[0]) / diff == mid)
low = mid + 1;
else
// the missing element will exist in left half
high = mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return arr[high] + diff;
}

// The function uses findMissingUtil() to
// find the missing element in AP.
// It assumes that there is exactly one
// missing element and may give incorrect result
// when there is no missing element or
// more than one missing elements. This function
// also assumes that the difference in AP is an
// integer.
int findMissing(int arr[], int n)
{
// If exactly one element is missing, then we can find
// difference of arithmetic progression using following
// formula. Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the difference is
// an integer.
int diff = (arr[n - 1] - arr[0]) / n;

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}
};

// Driver Code
int main()
{
GFG g;
int arr[] = {2, 4, 8, 10, 12, 14};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "The missing element is "
<< g.findMissing(arr, n);
return 0;
}

// This code is contributed by gurudev620gs
```

## C

```// A C program to find the missing number in a given
// arithmetic progression
#include <stdio.h>
#include <limits.h>

// A binary search based function that returns
// the missing element in arithmetic progression
int findMissingUtil(int arr[], int low, int high, int diff)
{
// Find index of middle element
int mid;
while (low <= high)
{
// find index of middle element
mid = (low + high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((arr[mid] - arr[0]) / diff == mid)
low = mid + 1;
else
// the missing element will exist in left half
high = mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return arr[high] + diff;
}

// The function uses findMissingUtil() to find the missing
// element in AP.  It assumes that there is exactly one missing
// element and may give incorrect result when there is no missing
// element or more than one missing elements.
// This function also assumes that the difference in AP is an
// integer.
int findMissing(int arr[], int n)
{
// If exactly one element is missing, then we can find
// difference of arithmetic progression using following
// formula.  Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the difference is
// an integer.
int diff = (arr[n-1] - arr[0])/n;

// Binary search for the missing number using above
// calculated diff
return findMissingUtil(arr, 0, n-1, diff);
}

/* Driver program to check above functions */
int main()
{
int arr[] = {2, 4, 8, 10, 12, 14};
int n = sizeof(arr)/sizeof(arr[0]);
printf("The missing element is %d", findMissing(arr, n));
return 0;
}
```

## Java

```
// A Java program to find
// the missing number in
// a given arithmetic
// progression
import java.io.*;

class GFG
{

// A binary search function that
// returns the missing
// element in arithmetic
// progression
static int findMissingUtil(int arr[], int low,
int high, int diff)
{
// Find index of middle element
int mid;
while (low <= high)
{
// find index of middle element
mid = (low + high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((arr[mid] - arr[0]) / diff == mid)
low = mid + 1;
else
// the missing element will exist in left half
high = mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return arr[high] + diff;
}

// The function uses findMissingUtil()
// to find the missing element in AP.
// It assumes that there is exactly
// one missing element and may give
// incorrect result when there is no
// missing element or more than one
// missing elements. This function also
// assumes that the difference in AP is
// an integer.
static int findMissing(int arr[], int n)
{
// If exactly one element is missing,
// then we can find difference of
// arithmetic progression using
// following formula. Example, 2, 4,
// 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that
// the difference is an integer.
int diff = (arr[n - 1] - arr[0]) / n;

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}

// Driver Code
public static void main (String[] args)
{
int arr[] = {2, 4, 8, 10, 12, 14};
int n = arr.length;
System.out.println("The missing element is "+
findMissing(arr, n));
}
}

// This code is contributed by gurudev620gs.

```

## Python3

```# A Python3 program to find the missing
# number in a given arithmetic progression
import sys

# A binary search based function
# that returns the missing element in
# arithmetic progression
def findMissingUtil(arr, low, high, diff):

while low <= high:
# find index of middle element
mid = (low + high)//2
# if mid == (nth position of element in AP)-1
# the missing element will exist in right half
if (arr[mid] - arr[0])//diff == mid:
low = mid + 1
else:
# the missing element will exist in left half
high = mid - 1
# after breaking out of binary search loop
# our missing element will exist between high and low
# our missing element will be a[high] + commom difference
# or a[low] - commom difference
return arr[high] + diff

# The function uses findMissingUtil() to find
# the missing element in AP. It assumes that
# there is exactly one missing element and may
# give incorrect result when there is no missing
# element or more than one missing elements.
# This function also assumes that the difference
# in AP is an integer.
def findMissing(arr, n):

# If exactly one element is missing, then
# we can find difference of arithmetic
# progression using following formula.
# Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
# The assumption in formula is that the
# difference is an integer.
diff = int((arr[n - 1] - arr[0]) / n);

# Binary search for the missing number
# using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);

# Driver Code
arr = [2, 4, 8, 10, 12, 14];
n = len(arr);
print("The missing element is",
findMissing(arr, n));

# This code is contributed by gurudev620gs

```

## C#

```// A C# program to find
// the missing number in
// a given arithmetic
// progression
using System;

class GFG
{

// A binary search function that
// returns the missing
// element in arithmetic
// progression
static int findMissingUtil(int []arr, int low,
int high, int diff)
{
// Find index of middle element
int mid;
while (low <= high)
{
// find index of middle element
mid = (low + high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((arr[mid] - arr[0]) / diff == mid)
low = mid + 1;
else
// the missing element will exist in left half
high = mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return arr[high] + diff;
}

// The function uses findMissingUtil()
// to find the missing element
// in AP. It assumes that there
// is exactly one missing element
// and may give incorrect result
// when there is no missing element
// or more than one missing elements.
// This function also assumes that
// the difference in AP is an integer.
static int findMissing(int []arr, int n)
{
// If exactly one element
// is missing, then we can
// find difference of arithmetic
// progression using following
// formula. Example, 2, 4, 6, 10,
// diff = (10-2)/4 = 2.The assumption
// in formula is that the difference
// is an integer.
int diff = (arr[n - 1] -
arr[0]) / n;

// Binary search for the
// missing number using
// above calculated diff
return findMissingUtil(arr, 0,
n - 1, diff);
}

// Driver Code
public static void Main ()
{
int []arr = {2, 4, 8,
10, 12, 14};
int n = arr.Length;
Console.WriteLine("The missing element is "+
findMissing(arr, n));
}
}

// This code is contributed by gurudev620gs.

```

## PHP

```<?php
// A PHP program to find the missing
// number in a given arithmetic progression

// A binary search based function
// that returns the missing element in
// arithmetic progression

function findMissingUtil(\$arr, \$low, \$high, \$diff)
{
while (\$low <= \$high)
{
// find index of middle element
\$mid = (\$low + \$high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((\$arr[\$mid] - \$arr[0]) / \$diff == \$mid)
\$low = \$mid + 1;
else
// the missing element will exist in left half
\$high = \$mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return \$arr[\$high] + \$diff;
}

// The function uses findMissingUtil() to find
// the missing element in AP. It assumes that
// there is exactly one missing element and may
// give incorrect result when there is no missing
// element or more than one missing elements.
// This function also assumes that the difference
// in AP is an integer.
function findMissing(\$arr, \$n)
{
// If exactly one element is missing, then
// we can find difference of arithmetic
// progression using following formula.
// Example, 2, 4, 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that the
// difference is an integer.
\$diff = (\$arr[\$n - 1] - \$arr[0]) / \$n;

// Binary search for the missing number
//  using above calculated diff
return findMissingUtil(\$arr, 0, \$n - 1, \$diff);
}

// Driver Code
\$arr = array(2, 4, 8, 10, 12, 14);
\$n = sizeof(\$arr);
echo "The missing element is ",
findMissing(\$arr, \$n);

// This code is contributed by gurudev620.gs
?>

```

## Javascript

```<script>

// A JavaScript program to find
// the missing number in
// a given arithmetic
// progression

// A binary search function that
// returns the missing
// element in arithmetic
// progression
function findMissingUtil(arr, low, high, diff)
{
// Find index of middle element
int mid;
while (low <= high)
{
// find index of middle element
mid = (low + high) / 2;
// if mid == (nth position of element in AP)-1
// the missing element will exist in right half
if ((arr[mid] - arr[0]) / diff == mid)
low = mid + 1;
else
// the missing element will exist in left half
high = mid - 1;
}
// after breaking out of binary search loop
// our missing element will exist between high and low
// our missing element will be a[high] + commom difference
// or a[low] - commom difference
return arr[high] + diff;
}

// The function uses findMissingUtil()
// to find the missing element in AP.
// It assumes that there is exactly
// one missing element and may give
// incorrect result when there is no
// missing element or more than one
// missing elements. This function also
// assumes that the difference in AP is
// an integer.
function findMissing(arr, n)
{
// If exactly one element is missing,
// then we can find difference of
// arithmetic progression using
// following formula. Example, 2, 4,
// 6, 10, diff = (10-2)/4 = 2.
// The assumption in formula is that
// the difference is an integer.
let diff = parseInt((arr[n - 1] - arr[0]) / n, 10);

// Binary search for the missing
// number using above calculated diff
return findMissingUtil(arr, 0, n - 1, diff);
}

let arr = [2, 4, 8, 10, 12, 14];
let n = arr.length;
document.write("The missing element is "+
findMissing(arr, n));

</script>```
Output
`The missing element is 6`

Time Complexity: O(log n)

Auxiliary Space: O(1)

Thanks to gurudev620gs for suggesting the above solution.

Exercise:
Solve the same problem for Geometrical Series. What is the time complexity of your solution? What about Fibonacci Series?