B + 2 D + 2 F + 3 I + 3 L + 4 P + 4 T

Each term consists of consecutive letters in order. The number of letters in the terms goes on increasing by one at each step.
Also, there is a gap of one letter between the last letter of the first term and the first letter of the second term; a gap of two letters between the last letter of the second term and the first letter of the third term; and so on.
So, there should be a gap of three letters between the last letter of the third term and the first letter of the desired term.

1st letter : A + 2 = C + 3 = F + 4 = J
2nd letter : Z - 2 = X - 3 = U - 4 = J

1st letter :A+1 = B, B+2 = D, D+3 = G, G+4 = K
2nd letter :Y-3 = V, V-4 = R, R-5 = M, M-6 = G
3rd letter :D+2 = F, F+2 = H, H+2 = J, J+2 = L

1st letter : D + 4 = H + 5 = M + 6 = S
2nd letter : E + 4 = I + 5 = N + 6 = T
3rd letter : F + 4 = J + 5 = O + 6 = U

1st letter : U - 2 = S - 4 = O - 2 = M - 4 = I
2nd letter : P - 8 = H - 4 = D - 2 = B - 1 = A
3rd letter : I + 1 = J + 6 = P + 1 = Q + 6 = W

In the first step, one letter from the beginning and one from the end of a term are removed to give the next term. In the second step, two letters from the beginning of a term are removed.
These two steps are repeated alternately.

R + 3 = U
U + 3 = X
X + 3 = A
A + 3 = D
D + 3 = G

1st letter : A + 6 = G + 6 = M + 6 = S + 6 = Y
2nd letter :Z - 6 = T - 6 = N - 6 = H - 6 = B

1st letter : G + 3 = J + 4 = N + 5 = S + 6 = Y + 7 = F
2nd letter : H + 4 = L + 5 = Q + 6 = W + 7 = D + 8 = L

1st letter :A + 1 B + 1 C + 1 D
2nd letter : I + 1 J + 1 K + 1 L

1st letter :B + 2 = D + 2 = F + 2 = H + 2 = J
2nd letter :Z -1 = Y - 1 = X - 1 = W - 1 = V
3rd letter :A + 2 = C + 2 = E + 2 = G + 2 = I

The given sequence is a combination of two series:
I. M, O, R, V and II. N, L, I, ?

We have 3 patterns:
I: A B C D ... X Y Z, which occurs only once.
Y X ... B A, which repeats alternately.
B C ... Y Z, which repeats alternately.
Now, I: has 26 terms.
So, number of terms before the desired term = (999 - 26) = 973.
Each of the patterns which occurs after I: has 25 letters.
Now, 973 ÷ 25 gives quotient = 38 and remainder = 23.
Thus, the 1000th trem of the given series is the 24th term of the 39th pattern after I:.
Clearly, the 39th pattern is II: and its 24th term is B.

Z - 6 T - 6 N - 6 H - 6 B
Z -3 W - 3 T - 3 Q- 3 N - 3 K - 3 H - 3 E - 3 B