Alphabet series with alternating large jumps – find the next letter: M, N, O, L, R, I, V, ?

Introduction / Context:
Some letter series mix short local moves with alternating large jumps. The goal is to detect the repeating sign pattern and how the jump sizes evolve. Here, the sequence after the initial M, N, O takes alternating forward and backward leaps whose magnitudes increase in a simple progression.

Given Data / Assumptions:

• Letters: M(13), N(14), O(15), L(12), R(18), I(9), V(22), ?
• Exactly one next term is asked.

Concept / Approach:
Compute successive differences (in positions) and observe the alternating signs and growing magnitudes. After the initial +1, +1, the pattern becomes “−, +, −, + …” with absolute jump sizes increasing by small, regular amounts.

Step-by-Step Solution:
M → N: +1; N → O: +1.O → L: −3; L → R: +6; R → I: −9; I → V: +13.Absolute jump sizes: 3, 6, 9, 13. The first three grow by +3 each (3, 6, 9). The next increase is +4, giving 13. Continuing the same gentle growth suggests the next magnitude increases by +4 again (13 + 4 = 17) while the sign alternates back to negative.V(22) − 17 = 5 = E. Therefore, the next letter is E.

Verification / Alternative check:
Projecting the pattern further (+? then −?) shows a smooth scheme of alternating signs with magnitudes 3, 6, 9, 13, 17… (growing by +3, +3, +4, +4). Plugging E matches this logic and maintains alternation.

Why Other Options Are Wrong:

• A, F, H, Z correspond to different jump sizes that break the gentle increase pattern or the sign alternation.

Common Pitfalls:

• Expecting constant steps or only +/−3k multiples, and missing that after three +3 increments the growth itself increases by 1.

Final Answer:
E