Alphanumeric Series

In this series, three patterns work together. Letters at the start go backwards by 2 (Z-X-V-T-R-P-N). The middle numbers follow the pattern: multiply by position and add the same digit (1×1+1=2, 2×2+2=6, 6×3+3=21, 21×4+4=88, 88×5+5=445, so 445×6+6=2676). Letters at the end move forward by 3 (A-D-G-J-M-P-S). Combining these gives N2676S.

This series has three interlinked patterns. First number: add the last number to get the next (2+5=7, 7+7=14, 14+9=23, 23+11=34, 34+13=47). Middle letters: go backwards one at a time (Z-Y-X-W-V-U). Last number: consecutive odd numbers (5, 7, 9, 11, 13, 15). This gives 47U15.

Three patterns operate here. First letters: +1, -2, +3, -4 (J-K-I-L-H-M). Middle numbers: consecutive differences increasing by 1 (2, 4, 7, 11, 16, 22 with differences of 2, 3, 4, 5, 6). Last letters: go backwards by 2 each time (Z-X-V-T-R-P). The missing term is L11T.

This follows a simple alphabetical and numerical pattern. First letters go forward by 2 (D-F-H-J-L-N). Numbers are consecutive even numbers (4, 6, 8, 10, 12, 14). The missing terms are L-12 and N-14, which match option C.

Three patterns work together. First number: multiply by increasing integers (2×2=4, 4×3=12, 12×4=48). Middle letters: advance by 3 (A-D-G-J). Last number: add consecutive even numbers (11+2=13, 13+4=17, 17+6=23). The missing term is 48J23.

Simple patterns govern this series. First letters: forward by 3 (C-F-I-L). Middle numbers: perfect squares (2²=4, 3²=9, 4²=16, 5²=25). Last letters: backward by 3 (X-U-R-O). The missing term is L25O.

The pattern here: first number increases by consecutive odd numbers (3+3=6, 6+5=11, 11+7=18, 18+9=27). The letter position increases by consecutive numbers (F+1=G, G+2=I, I+3=L, L+4=P). The answer is 27P.

Three patterns: first letter goes back by 2 (K-I-G-E-C), second letter goes forward by 3 (M-P-S-V-Y), number adds 3 each time (5, 8, 11, 14, 17). The missing term is CY17.

Letters go backwards by 2 (W-U-S-Q-O) and numbers are descending perfect squares (12²=144, 11²=121, 10²=100, 9²=81, 8²=64). The missing term is U-121.

Two interlinked patterns: numbers increase by consecutive even numbers (2+2=4, 4+4=8, 8+6=14, 14+8=22) and letters advance by increasing steps (B+1=C, C+2=E, E+3=H, H+4=L). The answer is 22L.

This alternating series has two patterns: numbers go +7, -3, +7, -3 (2+7=9, 9-3=6, 6+7=13, 13-3=10) while letters simply advance one at a time (A-B-C-D). The missing number is 10.

The series has three interwoven patterns. The first letters decrease by 3 (N→K→H→E→B). The numbers follow +2, +3, +4, +5 pattern (5→7→10→14→19). The last letters decrease by 2 (V→T→R→P→N). Only H10R satisfies all three patterns.

The first letters increase by 2 (Q→S→U→W→Y). The last letters decrease by 1 (F→E→D→C→B). The numbers follow n×(n+1)+(n+1): 1, 2, 6, 21, 88. Only Y88B satisfies all three conditions.

In 1CV, 5FU, 9IT, 15LS, 17OR, the first letters follow the sequence 1, 5, 9, 15, 17 (odd numbers but 17 should be 19). The last letters are V, U, T, S, R (backward consecutive). The middle letters follow C→F→I→L→O (+3 each). 15LS breaks the number pattern.

The first letters decrease by 2 each time (K→I→G→E→C). The middle numbers increase by 3 (5→8→11→14→17). The last letters decrease by 2 (M→P→S→V→Y). Only CY17 fits all three patterns.

The first letters increase by 3 (C→F→I→L). The middle numbers follow squares: 4, 9, 16, 25. The last letters decrease by 3 (X→U→R→O). L25O is the only option satisfying all patterns.

The series follows three patterns: first numbers are 2, 7, 14, 23, 34, 47 (differences 5, 7, 9, 11, 13 - consecutive odd numbers). Letters decrease by 1 (Z→Y→X→W→V→U). Last numbers increase by 2 (5, 7, 9, 11, 13, 15). Only 47U15 satisfies all.

In each step,both the numbers and letters are taking a forward step of 2. hence the next two letter after J will be L and N and the next two numbers after 10 are 12 and 14.

The first numbers follow n! pattern: 2, 4, 12, 48 (2!=2, 2!×2=4, 4!×3=12, 12×4=48). First letters increase by 2 (A→D→G→J). Last numbers increase by 2 (11→13→17→23). Only 48J23 satisfies all conditions.