Let the sum of the first three terms of an A.P. be 39 and the sum of its last four terms be 178. If the first term of this A.P. is 10, then the median of the A.P. is:

Let the first three terms of the A.P. be a-d, a, a+d
Given a-d=10.. (1)
Also given a-d+a+a+d=39 => 3a=39.. (2)
From (2), a=13
Substituting in (1), 13-d=10 => d=3
Let the A.P. have n terms. Then the last four terms are:
a+(n-4)d, a+(n-3)d, a+(n-2)d, a+(n-1)d
Their sum is given as 178.
Therefore, 4a + (n-4+n-3+n-2+n-1)d = 178
4a + (4n-10)d = 178
Substituting a=13 and d=3,
4(13) + (4n-10)(3) = 178
52 + 12n - 30 = 178
12n = 156
n = 13
The median is the middle term, which is the (n+1)/2 th term when n is odd.
In this case, n=13, so the median term is the 7th term.
The 7th term is a + 6d = 13 + 6(3) = 13+18 = 31
Therefore, the median of the A.P. is 31.