Let a=n, b=n+1, c=n+2 and d=n+3
1.(a) n+n+1+n+2+n+3=130
=>4n+6=130
=>4n=124
=>n=31, four consecutive numbers=31,32,33,34
(b) n+n+1+n+2+n+3=-38
=>4n-6=-38 =>4n=-44=>n=-11
Four consecutive numbers=-11,-10,-9,-8
2. Algebraic representation=n+(n+1)+(n+2)=(n+3)+10
=>3n+3=n+13 => 2n=10 => n=5
The numbers are:- 5,6,7,8 Proof:-5+6+7=18. 18-10=8
3. Take the pair of four consecutive numbers:- 5,6,7,8 and 68,69,70,71.
i. 5=n, 6=n+1,7=n+2 and 8=n+3
(8+5)+(6+7)=13-13=0
ii. 68=n, 69=n+1, 70=n+2, 71=n+3
(71+68)-(69+70)=139-139=0
Explanation:- ((n+3)+n)-((n+2)+(n+1))
=>(2n+3)-(2n+3)=0
4. Take the previous pair of four consecutive numbers 5,6,7,8 and 68,69,70,71.
i. 5+6+7=18. 18-8=10. 5x2=10
ii. 68+69+70=207.207-71=136. 68x2=136
The above examples demonstrate the fact that a+b+c-d=2a. As long as a,b,c and d are four consecutive numbers.
Proof:-
let a=n, b=n+1, c=n+2 and d=n+3
(n)+(n+1)+(n+2)-(n+3)=3n+3-n+3=2n
- Which proves that the sum of three of the four consecutive numbers minus the fourth equals to the first digit being doubled.