Q12/119) Let a < b < c < d be four real numbers, such that all six pairwise sums ...? are distinct. The values of the smallest four pairwise sums are 1, 2, 3, and 4 respectively. What are the possible values of d?

Then the six pairwise sums

a+b, a+c, a+d, b+c, b+d, c+d

are all distinct.
there are 2 possibilities
case 1:
a+d < b+ c

Now, a+b = 1, a+c = 2, a+d = 3, b+c = 4.

∴ (a+c) - (a+b) = 2-1

∴ c-b = 1

∴ (b+c) + (c-b) = 4+1 ∴ 2c = 5 ∴ c = 5/2

∴ a+c = 2 gives a = 2-c = 2-(5/2) = -1/2

∴ a+d = 3 gives d = 3-a = 3-(-1/2) = 7/2

∴ d = 7/2. ........ Ans for case 1.
 case2 : b+ c< a + d
then we get a+b = 1, a+c = 2, b+c =3 a +d = 4 => 2a + b + c = 3 => a = 0 b = 1 c = 2 and d = 4

so d =4 is another solution for case 2

so d can be 7/2 or 4