There is no solution to this problem.
Lets see the explanation.
The given sequence is in A.P with initial term, a=2 and common difference, d=4.
So the numbers are 2+4n1, 2+ 4n2, 2+4n3, and so on.
Now we need to add any 5 nos. in order to get total as 100.
Lets say these nos are 2+4n1, 2+4n2, 2+4n3, 2+4n4, 2+4n5.( Or for that matter u can take randomly as n7, n3, etc)
Now lets add these nos. and as we now sum should be 100,
Therefore, we have
(2+4n1) + (2+4n2) + (2+4n3) + (2+4n4) + (2+4n5) = 100
4( n1 + n2 + n3 + n4 + n5 ) + 10 = 100
4( n1 + n2 + n3 + n4 + n5 ) = 90
n1 + n2 + n3 + n4 + n5 = 90/4
Since 90 is not completely divisible by 4, we can conclude that there is no solution to the above problem.
Hence the proof.