Discover the Lacking Time period in an Arithmetic Development in Javascript

The problem

An Arithmetic Development is outlined as one in which there’s a continuing distinction between the consecutive phrases of a given sequence of numbers. You’re supplied with consecutive parts of an Arithmetic Development. There may be nonetheless one hitch: precisely one time period from the unique sequence is lacking from the set of numbers which were given to you. The remainder of the given sequence is identical as the unique AP. Discover the lacking time period.

It’s important to write a perform that receives an inventory, record dimension will all the time be no less than 3 numbers. The lacking time period won’t ever be the primary or final one.

Instance:

findMissing([1, 3, 5, 9, 11]) == 7

The answer in Javascript

Choice 1:

var findMissing = perform (l) {  
  return ((l[0]+l[l.length-1])*(l.size+1))/2-(l.scale back((a,b)=>a+b))
}

Choice 2:

var findMissing = perform (record) {
  let delta = (record[list.length - 1] - record[0])/record.size;
  return record.discover( (el,ind) => el !== ind*delta + record[0]) - delta;
}

Choice 3:

var findMissing = perform (record) {  
  var listLen = record.size;
  var iter = (record[listLen-1]-list[0])/listLen;
  
  for(i = 0; i < listLen; i++){
    var subsequent = record[i] + iter
    if(subsequent != record[i+1]){
      return subsequent;
    }
  }
}

Check instances to validate our answer

for (let i = 0;i < 10;++i) {
  fixture.random (1+i*5,0) ;
  fixture.random (1+i*5,10) ;
  fixture.random (1+i*5,100) ;
  fixture.random (1+i*5,1000) ;
  fixture.random (1+i*5,10000) ;
  fixture.random (1+i*5,100000) ;
}