Question 1

Alphabet series puzzle with positional increments: complete the pattern SCD, TEF, UGH, ____, WKL by analyzing letter-wise progressions and consistent offsets

Accepted Answer

Given data Series: SCD, TEF, UGH, ?, WKL. Concept/Approach (why this method) Handle each position independently. Look for constant alphabetical increments (A=1, B=2, …, Z=26) across first, second, and third letters. Step-by-Step calculation 1) First letters: S(19), T(20), U(21), ?, W(23) ⇒ +1 each step ⇒ missing is V(22).2) Second letters: C(3), E(5), G(7), ?, K(11) ⇒ +2 each ⇒ missing is I(9).3) Third letters: D(4), F(6), H(8), ?, L(12) ⇒ +2 each ⇒ missing is J(10).4) Combine: V + I + J ⇒ VIJ. Verification/Alternative Check neighbors: UGH → VIJ (U→V +1, G→I +2, H→J +2); VIJ → WKL (V→W +1, I→K +2, J→L +2) — pattern holds. Common pitfalls Applying a single uniform step to all positions; here the first letter uses +1 while the other two use +2. Final Answer VIJ

Question 2

Alpha–numeric series completion: identify the missing term in B2CD, _____, BCD4, B5CD, BC6D by tracking digit progression and shifting digit position

Accepted Answer

Given data Series: B2CD, _____, BCD4, B5CD, BC6D. Concept/Approach (why this method) Two patterns run simultaneously: (i) the number increases by 1 each step (2,3,4,5,6); (ii) the position of the number cycles through the 2nd→3rd→4th characters, then repeats. Step-by-Step calculation 1) Term 1: B2CD (number in 2nd position).2) Term 2 (missing): number should be 3 and shift to 3rd position ⇒ BC3D.3) Term 3: BCD4 (4th position).4) Term 4: B5CD (cycle repeats to 2nd position).5) Term 5: BC6D (3rd position). Verification/Alternative Positions follow the loop (2→3→4→2→3), and the digits are 2,3,4,5,6 in order — consistent. Common pitfalls Placing the '3' in the second position, breaking the positional cycle. Final Answer BC3D

Question 3

Letter pattern with alternating swaps: complete the sequence FAG, GAF, HAI, IAH, ____ by recognizing swap/no-swap alternation and +1 increments across first and third letters

Accepted Answer

Given data Series: FAG, GAF, HAI, IAH, ____. Concept/Approach (why this method) Observe two simultaneous rules: (i) the first and third letters each increase by +1 per step overall (net +2 every two terms), and (ii) terms alternate between 'normal order' and 'swapped' order of the first/third letters. Step-by-Step calculation 1) T1 = F A G (normal order).2) T2 = G A F (swap of T1's 1st and 3rd).3) T3 = H A I (both letters +2 from T1: F→H, G→I; normal order).4) T4 = I A H (swap of T3).5) T5 must be normal order again with +2 from T3: H→J and I→K ⇒ J A K. Verification/Alternative Check increments: (F,G)→(H,I)→(J,K) in normal terms; and each alternate term swaps those two letters, matching observed T2 and T4. Common pitfalls Applying +1 rather than the two-step +2 across non-adjacent normal terms. Forgetting the swap/no-swap alternation. Final Answer JAK

Question 4

Alphabetical series reasoning puzzle: complete the sequence ELFA, GLHA, ILJA, ____, MLNA by identifying the letterwise patterns (first and third letters move by +2 in the alphabet while the second and fourth positions remain fixed) to find the next …

Accepted Answer

We analyze each character position across the given terms. Positions by term: ELFA, GLHA, ILJA, ? , MLNA. 1st letter: E → G → I → (add +2 each step) → K → M. 2nd letter: constant 'L' throughout. 3rd letter: F → H → J → (add +2 each step) → L → N. 4th letter: constant 'A' throughout. Concept/Approach Treat each column independently. Detect constant columns and arithmetic shifts (+2) in alphabetic indices for the others. Step-by-Step calculation 1) First column: E(5) + 2 = G(7), +2 = I(9), +2 = K(11), +2 = M(13).2) Second column: always L ⇒ L.3) Third column: F(6) + 2 = H(8), +2 = J(10), +2 = L(12), +2 = N(14).4) Fourth column: always A ⇒ A. Verification/Alternative Rebuild the missing term from columns: K + L + L + A = KLLA. It fits neatly between ILJA and MLNA. Common pitfalls Mixing the +1 and +2 strides; overlooking that the 2nd and 4th characters are fixed constants. Final Answer KLLA

Question 5

Complete the pattern in the series CMM, EOO, GQQ, ____, KUU by tracking three independent alphabetic progressions (each moving forward by +2 letters) to determine the missing triple.

Accepted Answer

Observe three synchronized +2 sequences. 1st letters: C → E → G → ( +2 ) → I → K. 2nd letters: M → O → Q → ( +2 ) → S → U. 3rd letters: M → O → Q → ( +2 ) → S → U. Concept/Approach Treat each position as an arithmetic progression on alphabet indices: A=1, B=2, …, Z=26. Step-by-Step calculation 1) First column: C(3)+2=E(5), +2=G(7), +2=I(9).2) Second column: M(13)+2=O(15), +2=Q(17), +2=S(19).3) Third column: same as second ⇒ next is S. Verification/Alternative Compose the missing item: I + S + S = ISS. The final given term KUU continues the +2 stride. Common pitfalls Accidentally stepping by +1; confusing the mirrored second and third positions. Final Answer ISS

Question 6

Alpha–numeric series completion: determine the next term after ZA5, Y4B, XC6, W3D by decoding the simultaneous letter shifts (first letter −1, last letter +1) and an alternating numeric pattern.

Accepted Answer

There are three independent streams: first letter descends, number alternates −1/+2/−3/+4, last letter ascends. First letters: Z → Y → X → W → V (−1 each step). Numbers: 5 → 4 (−1), 4 → 6 (+2), 6 → 3 (−3), 3 → 7 (+4) — continuing the alternating magnitude pattern. Last letters: A → B → C → D → E (+1 each step). Concept/Approach Decompose the string into (Letter1, Number, Letter2) and analyze each as its own progression. Step-by-Step calculation 1) Letter1: Z(26)−1 → Y(25)−1 → X(24)−1 → W(23)−1 → V(22).2) Number: 5 −1 = 4; 4 +2 = 6; 6 −3 = 3; 3 +4 = 7.3) Letter2: A(1)+1 → B(2)+1 → C(3)+1 → D(4)+1 → E(5). Verification/Alternative Combine results: V + 7 + E = V7E. Common pitfalls Missing the alternating sign and increasing magnitude in the number sequence; assuming a constant numeric step. Final Answer V7E

Question 7

Reverse alphabet triplets puzzle: fill the blank in QPO, NML, KJI, ____, EDC by continuing groups of three consecutive letters in descending order with a −3 shift on the starting letter.

Accepted Answer

Each term is a three-letter block in strict reverse order; the block's first letter descends by 3 between terms. Blocks: QPO (Q→P→O), NML (N→M→L), KJI (K→J→I), ? , EDC (E→D→C). Start letters: Q → N → K → H → E (−3 each step). Concept/Approach Identify the window size (3 letters) and the stride (−1 inside windows, −3 between windows). Step-by-Step calculation 1) After KJI, subtract 3 from K ⇒ H; write H in reverse block: H, G, F ⇒ HGF.2) Next subtract 3 from H ⇒ E, yielding the given EDC. Verification/Alternative Sequence of start letters: Q(17), N(14), K(11), H(8), E(5) confirms −3 stride. Common pitfalls Proposing ascending JKL (opposite to the within-block direction) or miscounting the start-letter stride. Final Answer HGF

Question 8

Alphabet stepping pattern across three positions: continue JAK, KBL, LCM, MDN, ____ by noting +1 shifts in all three letter positions to produce the next code.

Accepted Answer

Every position advances by +1 letter from term to term. First letters: J → K → L → M → N (+1 each). Second letters: A → B → C → D → E (+1 each). Third letters: K → L → M → N → O (+1 each). Concept/Approach Treat it as three parallel arithmetic progressions on letters with unit step. Step-by-Step calculation Next term = (M+1, D+1, N+1) = (N, E, O) ⇒ NEO. Verification/Alternative Check adjacency: NEO precedes nothing given, but backward check confirms MDN → NEO is +1 in all positions. Common pitfalls Altering only one position or skipping letters due to misalignment. Final Answer NEO

Question 9

Complete the letter–mirror pattern BCB, DED, FGF, HIH, ___ by recognizing that each of the three positions advances by +2 letters, retaining the palindromic structure.

Accepted Answer

Palindromic triplets with uniform +2 steps across positions. First letters: B → D → F → H → J (+2). Middle letters: C → E → G → I → K (+2). Last letters mirror the first: B → D → F → H → J. Concept/Approach Identify symmetry (1st = 3rd) and apply a consistent +2 stride. Step-by-Step calculation Compute next letters: (H+2, I+2, H+2) = (J, K, J) ⇒ JKJ. Verification/Alternative Alphabet indices: B(2),D(4),F(6),H(8),J(10); C(3),E(5),G(7),I(9),K(11). Common pitfalls Breaking the palindrome or using a +1 step at any position. Final Answer JKJ

Question 10

Find the missing alphanumeric code in P5QR, P4QS, P3QT, ____, P1QV by tracking a countdown digit and an incrementing terminal letter while the initial letters stay fixed.

Accepted Answer

Structure: 'P' + digit + 'Q' + trailing letter. First letter: constant 'P'. Digit: 5, 4, 3, 2, 1 (counting down by 1). Third character: constant 'Q'. Last letter: R, S, T, U, V (+1 each step). Concept/Approach Decompose the token and apply independent progressions: numeric countdown; alphabetic +1 on the tail. Step-by-Step calculation Next after P3QT is P2QU with digit 2 and tail advanced from T to U. Verification/Alternative Final given term P1QV confirms the countdown to 1 and tail letter V. Common pitfalls Incrementing the digit instead of decrementing or shifting the wrong letter (the Q is fixed). Final Answer P2QU

Question 11

Alphabetic sequence with fixed middle: extend QAR, RAS, SAT, TAU, ____ by advancing the first and last letters by +1 while keeping the central 'A' unchanged.

Accepted Answer

The form is (Letter, 'A', Letter) with +1 steps on the outer letters. First letters: Q → R → S → T → U (+1). Second letters: constant 'A'. Third letters: R → S → T → U → V (+1). Concept/Approach Hold the middle fixed; treat the ends as synchronized forward progressions. Step-by-Step calculation After TAU, increment first and last: T→U, U→V ⇒ UAV. Verification/Alternative Check neighbors: SAT → TAU → UAV confirms steady +1 on the ends. Common pitfalls Altering the middle 'A' or reversing the direction on one side. Final Answer UAV

Question 12

Pattern of escalating exponents on D, E, F: complete DEF, DEF2, DE2F2, ____, D2E2F3 by identifying the leftward sweep of exponent increments to fill in the missing fourth term.

Accepted Answer

View each string as D^a E^b F^c, where omitted exponents equal 1. Exponents increase by +1 one letter at a time, cycling right-to-left. Term 1: D1E1F1 (shown as DEF). Term 2: D1E1F2 (increment F). Term 3: D1E2F2 (increment E). Term 4: D2E2F2 (increment D) ⇒ missing term. Term 5: D2E2F3 (increment F again) — given. Concept/Approach Recognize the incremental sweep of exponents: F ↑, then E ↑, then D ↑, then F ↑ again, … Step-by-Step calculation 1) After DE2F2, increment D's exponent: 1 → 2.2) Keep E and F at 2: compose D2E2F2. Verification/Alternative The provided last term D2E2F3 confirms the cycle restarting at F. Common pitfalls Jumping multiple exponents in one term or changing letter order. Final Answer D2E2F2

Question 13

Fill the blanks with one ordered 5-letter string to complete the series: cc_ccbc_accbcc_c_b_.

Accepted Answer

Introduction / Context: This verbal-reasoning item asks you to choose one 5-letter sequence that, when inserted left-to-right into the blanks, produces a coherent letter-run using the alphabet subset {a, b, c}. Such questions test pattern discovery, chunking of micro-blocks, and verification across all blanks. Given Data / Assumptions: Skeleton: c c _ c c b c _ a c c b c c _ c _ b _ Options (5 letters each) to be placed in order: acacc, abacc, ababc, aabcc. Goal: produce a smooth sequence with recurring sub-blocks (e.g., cc, cbc, acc, bcc) and no contradictions. Concept / Approach: The visible fragments suggest tiling of “cc”, “cbc”, and “acc/bcc” motifs. We test each candidate by inserting its letters in sequence and checking whether the completed string preserves these micro-blocks without creating stranded or clashing bigrams (like “ba” at awkward joints). Step-by-Step Solution: 1) Insert “abacc” at the blanks sequentially. The early region “cc_ab…” yields “ccab…”, which cleanly transitions into “…ccbc…”.2) Mid-string, the “a” and “b” placements bridge “…_accbcc…” into “…aaccbcc…”, preserving characteristic “acc” and “bcc” tiles used throughout.3) Later positions maintain the same texture, producing “c…b…” joints that match the skeleton’s alternation of b and c around stable cc cores.4) Alternative candidates introduce clashes: e.g., “ababc” creates repeated “bab” or “abc” at positions that disrupt the cc/cbc clusters; “aabcc” generates excessive “aac/abcc” clumps; “acacc” misaligns a critical mid transition. Verification / Alternative check: After insertion, re-scan the entire line to confirm that the sequence features recurring units like cc, cbc, acc, and bcc in positions the skeleton suggests. The “abacc” fill keeps these intact across all joints. Why Other Options Are Wrong: acacc: Creates awkward “cac” at a boundary where “cbc/acc” is expected. ababc: Produces “bab” collisions that break the c-centric texture. aabcc: Over-weights “a” near “cc” zones, distorting repeated cc-anchored tiles. Common Pitfalls: Checking only the first or last blank. You must place all five letters and verify every joint for coherence. Final Answer: abacc

Question 14

Choose the 4-letter string that completes the pattern: _stt_tt_tts_.

Accepted Answer

Introduction / Context: We must insert one ordered 4-letter option into the blanks to yield a consistent s/t pattern that repeats short motifs. The string features clusters “stt” and “tts” that hint at overlapping triplets. Given Data / Assumptions: Skeleton: _ s t t _ t t _ t t s _ Options: tsst, sstt, ttst, tsts (letters used in order). Concept / Approach: Look for a fill that supports rolling triplets like “s t t” and “t t s” without breaking adjacency when overlaps occur. The candidate should produce a smooth alternation of single s’s introducing or trailing blocks of t’s. Step-by-Step Solution: 1) Try “s s t t”. Place s at blank1: s s t t s… aligns two leading s’s into a natural preface before a “tt”.2) Continue: inserting s at blank2 preserves “…stt s tt…”.3) Insert t at blank3 to maintain “…tt t tt s…”.4) Insert t at blank4 to complete the tail “…tts t”, leaving well-formed “stt/tts” overlaps. Verification / Alternative check: Read the finished line with sliding windows of length 3 and note the recurrence of “stt” and its rotation “tts”. Why Other Options Are Wrong: tsst / ttst / tsts: Each creates at least one local inversion (e.g., “tst”) that breaks the intended triplet overlaps in the mid-segment. Common Pitfalls: Matching only the first blank; always confirm all four placements. Final Answer: sstt

Question 15

Insert one ordered 4-letter set to complete the pairwise pattern: ac_ga_eg_ce_.

Accepted Answer

Introduction / Context: The string shows two-letter pairs separated by underscores: ac, ga, eg, ce, with four blanks. The task is to select one 4-letter sequence to fill these blanks in order so that all adjacent pairs exhibit a consistent cyclic relation. Given Data / Assumptions: Visible pairs: a–c, g–a, e–g, c–e. Options to place in sequence: dbag, ecag, deag, ebdg. Concept / Approach: Observe that the right letter of one pair becomes the left letter of the next (…c → ga; a → eg; g → ce). This forms a cycle c → a → g → e → c. We therefore want the inserted letters to continue this cycle smoothly at each joint. Step-by-Step Solution: 1) The cycle c→a→g→e suggests linking letters d and e as neutral bridges while preserving the “a-c-g-e” loop.2) Testing “deag”: placing d, e, a, g in the four blanks yields joints that keep the c–a–g–e turns intact and align the mid-sequence overlaps.3) Alternative candidates create mismatched turns (e.g., “dbag” pushes b into positions that disrupt the c→a or a→g transitions). Verification / Alternative check: Read the finished line and confirm that consecutive bigrams rotate through the cycle c–a–g–e without contradiction. Why Other Options Are Wrong: dbag / ecag / ebdg: Each introduces at least one joint where the expected next pair is broken (e.g., c→b or a→c interruption). Common Pitfalls: Focusing only on individual pairs; you must track how each pair hands off to the next one. Final Answer: deag

Question 16

Fill with one ordered 4-letter string to complete the run: ac_bacc_a_cba_cb.

Accepted Answer

Introduction / Context: This sequence is built from the letters a, b, c with stable tiles like “ac”, “bacc”, and “cba”. We must insert one 4-letter option to realize a coherent concatenation of these tiles. Given Data / Assumptions: Skeleton: a c _ b a c c _ a _ c b a _ c b Options (4 letters each): cbcc, cbbc, caac, bbcb. Concept / Approach: We look for a fill that (i) closes “ac_ba” into “acb a…”, (ii) keeps “bacc” intact, and (iii) bridges into “…cba…” without creating clashes like “aa” or “bbb” where the pattern favors alternating around c. Step-by-Step Solution: 1) Try “cbbc”: insert c then b to obtain “ac c b a c c…”, which keeps “acc” and then “bacc” contiguous.2) The third letter b sustains the b-centered mid segment, and the fourth letter c prepares the handoff into “cba”.3) Alternatives introduce inconsistencies: “cbcc” yields over-stacked c’s; “caac” creates double a’s in a sensitive zone; “bbcb” leaves an early b-b congestion that breaks the “ac/bacc” cadence. Verification / Alternative check: Re-scan to see standard sub-blocks “ac”, “bacc”, “cba” appearing with clean joints. Why Other Options Are Wrong: cbcc: Too many consecutive c’s mid-run. caac: Produces “aa” misfit across an “ac/ca” boundary. bbcb: Forces b-clusters that mute the required c-anchors. Common Pitfalls: Greedy local fits that ruin later obligatory tiles like “cba”. Final Answer: cbbc

Question 17

Insert one ordered 5-letter string to complete the pattern: _c_ca_ab_bc_.

Accepted Answer

Introduction / Context: The skeleton shows overlapping bigrams “c_”, “_c”, “ca”, “ab”, “bc” from the alphabet subset {a, b, c}. We must pick a 5-letter fill that restores the intended repeating chain. Given Data / Assumptions: Skeleton: _ c _ c a _ a b _ b c _ (5 blanks) Options: abcac, babca, ccabb, acbba. Concept / Approach: We expect a cyclic run “…a→b→c→a→b→c…”. The best fill should yield recognizable overlaps like “ab”, “bc”, and “ca” at the designated joints without stutters. Step-by-Step Solution: 1) Insert “a b c a c” in order.2) Early blanks form “…a c c a…”, aligning into “…a c | c a…”, then “…a b…”, then “…b c…”, producing the target overlaps.3) Other options either double consonants at bad spots (“ccabb”), invert transitions (“acbba”), or misalign the mid “ab→bc” hinge (“babca”). Verification / Alternative check: Scanning with bigram windows shows “ca”, “ab”, and “bc” occurring in the right places. Why Other Options Are Wrong: babca: Misplaces the early transition and collides with the “c_” start. ccabb: Over-concentrates c’s and breaks the ab/bc alternation. acbba: Creates “bb” mid-sequence, which the skeleton does not support. Common Pitfalls: Not tracking how each blank affects the adjacent bigrams on both sides. Final Answer: abcac

Question 18

Fill with one ordered 5-letter string: _bcab_a_cab_a_c.

Accepted Answer

Introduction / Context: This is a straightforward completion task where only one option has the required length to fill all blanks. We must still verify that it produces smooth overlaps consistent with “bcab” and “cab” tiles. Given Data / Assumptions: Skeleton: _ b c a b _ a _ c a b _ a _ c Number of blanks: 5. Options lengths: acbcb (5), aabc (4), bbca (4), abac (4). Concept / Approach: Only “acbcb” can populate all five blanks. We still check adjacency to ensure it doesn’t break the visible “bcab / cab” motifs. Step-by-Step Solution: 1) Insert a, c, b, c, b into the five blanks left-to-right.2) The resulting joints produce “…a b c a b c a c a b c a b a c…”, which keeps “bcab” and “cab” intact at their anchor positions.3) Shorter options cannot fill all blanks; therefore they are invalid by format. Verification / Alternative check: Scan for any doubled letters that would contradict the skeleton’s alternation—none occur. Why Other Options Are Wrong: They have only 4 letters and cannot fill 5 blanks, violating the problem’s structure. Common Pitfalls: Not counting blanks before choosing. Final Answer: acbcb

Question 19

Complete the repeating block pattern by filling one 4-letter string: H_JH_IJHHI_HH_JH. Choose the set of four letters that, when placed in order, yields a consistent repeating sequence.

Accepted Answer

Introduction / Context: This problem features a letter sequence with missing letters in multiple positions. The aim is to identify a short repeating block and supply a single 4-letter fill that restores the repeated pattern across the entire string. Such items test pattern discovery and global consistency checks, not just local neighbor-matching. Given Data / Assumptions: Skeleton: H _ J H _ I J H H I _ H H _ J H We must choose one 4-letter sequence to replace the blanks in order (blank1 → blank2 → blank3 → blank4). Concept / Approach: Scan for a likely repeating unit. The visible chunks “H J H” and “I J H H I” suggest the motif HIJH or a near variant. If the complete string decomposes into HIJH blocks, each four-letter segment should read H, I, J, H cyclically. Step-by-Step Solution: 1) Try to enforce HIJH as the repeating block: H I J H | H I J H | H I J H | H I J H.2) Fill blank1 with I to get the first full block HIJH starting at position 1.3) Fill blank2 with H so that the next letters maintain the cycle H I J H | H…4) Fill blank3 with I so the mid-run becomes …H I J H H I J H… consistent with repeated HIJH.5) Fill blank4 with J to complete the final HIJH block at the tail.6) The four inserted letters in order are I, H, I, J → “IHIJ”. Verification / Alternative check: After insertion, segment the line into four-letter groups: HIJH | HIJH | HIJH | HIJH. Each block matches perfectly, confirming uniform periodicity. Why Other Options Are Wrong: IHJI / HIHI / HJHJ: Each produces at least one block with broken order or doubled letters at the wrong positions, destroying the clean HIJH repetition. Common Pitfalls: Solvers often fill blanks greedily (matching immediate neighbors) and miss the four-letter macro-cycle. Always hypothesize a block, then test the entire string for global fit. Final Answer: IHIJ

Question 20

Fill the gaps to complete the reverse–then-forward alphabet pattern: ZYX_W_YZZ_XWWXY_. Choose the 4-letter string that correctly fills all blanks in order.

Accepted Answer

Introduction / Context: Letter-series completion tasks measure a candidate’s ability to detect simultaneous local and global patterns in alphabetic runs. This item has a visible “descending then ascending” motif, with four blanks to be filled by one ordered 4-letter string chosen from the options. Given Data / Assumptions: Visible skeleton: Z Y X _ W _ Y Z Z _ X W W X Y _ English alphabet order; wrap-around is not required. One option (a 4-letter sequence) must fill the 4 blanks in order. Concept / Approach: The string begins with a strict reverse run Z, Y, X, then appears to pivot toward a forward movement (…W, X, Y, Z…). Many classic constructions mirror around a central turn. Hypothesis: the missing letters are precisely the forward block W, X, Y, Z to transition cleanly from the reverse segment into an organized forward cluster that repeats locally. Step-by-Step Solution: 1) Insert W at blank1 to complete the reverse tail: Z, Y, X, W.2) Insert X at blank2 to begin a forward push: …W, X, Y, Z…3) Insert Y at blank3 to continue the forward quartet near Z Z: …Y, Z, Z, X…4) Insert Z at blank4 so the tail reads …W, W, X, Y, Z with smooth local symmetry.Resulting string: Z Y X W W X Y Z Z X W W X Y Z. Verification / Alternative check: Read through the finished line: the early Z→Y→X descent passes into an overlapping forward cluster W→X→Y→Z that reappears as interlocking sub-blocks (…W X Y Z… and …X W W X Y Z…), which is typical of exam patterns that splice a clean forward run into a reverse-to-forward hinge. Why Other Options Are Wrong: WYXZ / WXZY / XYZW: Each scrambles the pure forward W→X→Y→Z order. Those permutations introduce local inversions (e.g., Y before X) that break the intended hinge from reverse to forward. Common Pitfalls: Many solvers focus on only the immediate neighbors of each blank and miss the overarching “reverse then forward” dual structure. Always test a candidate fill across all blanks and scan the entire line for clean runs. Final Answer: WXYZ

Letter and Symbol Series Questions