Let ABCDE be a Pentagon such that A=B=C=D=120°. All 5 sides are consecutive natural numbers but may not be in order i.e. sides can be 1,3,4,2,5 or 7,9,8,11,10 etc. (This is just example. These may be or may not be the possible solutions of sides of Pentagon.)
Define set S which contains all possible solutions of AB+BC+CD.
Find the sum of all elements present in S.
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Sum of angles in Pentagon is 540° so E=60°. Extend sides AB and CD to meet at F. Now in the quadrilateral AFDE, A=D=120° and E=60° so F=60° therefore the quadrilateral is a parallelogram. Also note that BCF is and equilateral triangle. From this, DE=AF=AB+BF and AE=DF=DC+CF and BC=CF=BF. So we get the following 2 equations:-
DE=AB+BC
AE=BC+CD
Let AB+BC+CD=P which we need to find
Add the equations to get P=AE+DE-BC
Let the sides be x,x+1...x+4 where x is a natural number. Obviously, AB,BC,CD can't be the longest side so either of AE or DE is x+4. Also AE,DE can't be the x,x+1 so if one of them is x+4 the other can be x+2,x+3. Also BC<DE,AE so BC can be x,x+1,x+2.
Case 1: when AE,DE are x+4,x+3
P=3x+3
BC=AE+DE-P=4-x
Now BC can be x,x+1,x+2 from the above explanation, putting those values of BC we get x=2,3/2,1 3/2 is not possible as sides are natural numbers. Therefore x=1,2 or P=6,9.
Case 2: when AE,DE are x+4,x+2
P=3x+4
BC=AE+DE-P=2-x
Now BC can be x,x+1 so x=1,1/2 but 1/2 is not possible so x=1 or P =7
Therefore P can be 6,7,9.
Hence, S={6,7,9} and the sum=6+7+9=22