Each term of the series is obtained by removing two letters from the preceding term one from the beginning and one from the end, So, the missing term is PENDICU.

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

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

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

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.

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

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

There is a gap of four letters between the first and second, the second and third letters of each term, and also between the last letter of a term and the first letter of the next term.

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.

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

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

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