A difficult maths question?

the odd terms are
u1 + u3 + u5..... u19
a + a + 2d + a+4d .... a + 18d
so
10 a + 90 d = 220
a + 9d = 22 (1)

even terms are

u2 + u4 + u6 .. + u20
a + d + a + 3d ... + a + 19d
so 10a + 100d = 250
a + 10d = 25 (2)

from (2) - (1)
d = 3
a = -5

therefore the two middle terms:
u10 = a + 9d= 22
u11= a + 10d = 25

notice that terms u10 and u 11 are equivalent to (1) and (2)- we needn't have calculated a and d

If it is an arithmetic sequence, then it means that each term increases by a constant step (let's call it k)
such that the first number is
a01 and the last number is a20
and the terms go like this

a01 = a01
a02 = a01 + k
a03 = a02 + k = a01 + 2k
and so on.

even terms added together:

a02 + a04 + a06...
(a01+k) + (a01+3k) + (a01+5k)+...
10(a01) + (1+3+5+7+9+11+13+15+17+19)k = 250

odd terms added together:
10(a01) + (0+2+4+6+8+10+12+14+16+18)k = 220

subtract one from the other
even - odd =
10(a01) - 10(a01) + (1+3+...+19)k - (0+2+...18)k = 250 - 220

(1+1+1+1+1+1+1+1+1+1)k = 30
10 k = 30
k = 3

Now we need to find a01. Use either one
(let's use odd)
10(a01) + (0+2+4+6+8+10+12+14+16+18)3 = 220
10(a01) + 90*3 = 220
10(a01) = 220 - 270 = -50
a01 = -5

Sequence must be:

-5, -2, +1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52