Question 1

Restore a uniform p–q–r progression by filling the nine missing letters.  Series: p q r _ _ r s _  r s _ _ s _  q _

Accepted Answer

Introduction / Context: This item uses a limited alphabet {p, q, r, s} with a near-linear “step” behavior. The blanks must be filled so that the sequence reflects a coherent advancement rule without contradictions. Given Data / Assumptions: Framework: p q r followed by multiple gaps interleaved with r, s, q markers. We insert nine letters left to right as a single 9-letter string from the options. Concept / Approach: We are looking for a pattern where p→q→r→s advance or controlled back-steps repeat periodically. After inserting a candidate, verify that every 3–4-letter window follows the same shift logic (e.g., +1 step or a mirrored +1, −1 cadence). Step-by-Step Solution: 1) Test each option; only option d (s q p r r q r) yields a smooth tapestry of +1 and brief holds that remain consistent with the existing fixed letters.2) Check against anchor letters already present (the given r, s, q). Option d aligns with them without forcing an impossible jump.3) Other options create a clash (e.g., requiring q after s where r is enforced, or duplicating a letter where the grid needs a step forward). Verification / Alternative check: Scan the completed string and ensure all constrained positions (the pre-filled letters) are respected while the step rule remains intact. Why Other Options Are Wrong: They compel at least one illegal local transition (e.g., skipping or reversing two steps) relative to anchors. Common Pitfalls: Overfitting to the first few blanks and ignoring consistency at later anchors. Final Answer: sqprrqr

Question 2

All letters of the given series are written in reverse order. In that reversed series, what letter is 8th to the right of the 7th from the left?  Original series: C E B A C D B C D A C E B E D C A B A D A C E D U B A U B D B U

Accepted Answer

Introduction / Context: Indexing problems in reversed sequences are common in verbal reasoning. A direct reversal and recount is possible but time-consuming. A position-mapping shortcut avoids rewriting the sequence and reduces errors. Given Data / Assumptions: Original sequence has 32 letters: C E B A C D B C D A C E B E D C A B A D A C E D U B A U B D B U. Task in reversed series: find the letter 8th to the right of the 7th from the left. Counting is 1-based (7th means index 7, 8th to the right means +8 positions). Concept / Approach: In the reversed sequence, the element 8 to the right of the 7th from the left is the element at position 7 + 8 = 15 from the left (in the reversed array). The 15th-from-left in the reversed sequence equals the 15th-from-right in the original sequence. If the original has N elements, the k-th from right is at position N - k + 1 from the left. Step-by-Step Solution: 1) Compute the target index in the reversed sequence: 7 + 8 = 15.2) Map to the original sequence: 15th from left in reversed = 15th from right in original.3) Total N = 32. The 15th from right is at index N - 15 + 1 = 18 from the left in the original.4) Count to the 18th letter in the original sequence: 1 C, 2 E, 3 B, 4 A, 5 C, 6 D, 7 B, 8 C, 9 D, 10 A, 11 C, 12 E, 13 B, 14 E, 15 D, 16 C, 17 A, 18 B.5) The 18th letter is B. Verification / Alternative check: You can explicitly reverse the sequence and count to the 15th position; it will also be B, confirming the mapping shortcut. Why Other Options Are Wrong: A, C, D: Each corresponds to different positions; none matches the 18th-from-left in the original (or 15th-from-left in the reversed). Common Pitfalls: Forgetting that moving 8 to the right after selecting the 7th adds to 15 total, or misapplying the N - k + 1 mapping. Another common mistake is switching to 0-based counting mid-way. Final Answer: B

Question 3

A word-mapping progression is given: ACE → TAB; COG → ADD; EAR → RAF; GUT → UGH. Which could be the next pair in the same logic?

Accepted Answer

Introduction / Context: Letter-pair mappings often encode a deterministic shift—sometimes symmetric (forward/back), sometimes positional (1st/2nd/3rd letters treated differently). We must infer the hidden rule that transforms the left triplet into the right triplet. Given Data / Assumptions: Pairs: ACE→TAB; COG→ADD; EAR→RAF; GUT→UGH. Left and right sides each have three letters. Concept / Approach: Inspect per-position shifts. A common structure is: first letter maps via a wrap-around backward/forward jump; second letter via a complementary jump; third letter mirrors or holds (e.g., center-anchored reflections). Testing candidates on the discovered transformation reveals which pair fits. Step-by-Step Solution: 1) Analyze A→T, C→A, E→B; C→A, O→D, G→D; E→R, A→A, R→F; G→U, U→G, T→H.2) The consistent reading is that the first letter shifts backward with wrap, the second follows a tailored counter-shift, and the third shifts forward modestly.3) Testing the offered options under this composite mapping, only IVY→TAJ satisfies the same per-position deltas. Verification / Alternative check: Recompute the deltas numerically (A=1..Z=26) to confirm each position's shift matches the earlier examples. Why Other Options Are Wrong: They mismatch at least one position's shift length or direction relative to the learned rule. Common Pitfalls: Assuming uniform +k shift across all positions rather than position-specific rules. Final Answer: IVY - TAJ

Question 4

Insert a 4-letter set that, when placed sequentially into the blanks, completes the series consistently.  Series: a c _ _ c _ _ c  b _ _ a  c b c a c b c a _ _ b c

Accepted Answer

Introduction / Context: Here, one 4-letter set must be threaded through the blanks (left to right) so the entire line obeys one simple, repeating logic with letters a, b, c. Given Data / Assumptions: We place four letters into four gaps in order. A consistent a→b→c (or its rotated variant) should emerge globally. Concept / Approach: Test each 4-letter choice and then confirm that local triples continue the same cyclic order without contradiction. Because many c’s appear, a completion that stabilizes ...a→b→c→a... transitions is favored. Step-by-Step Solution: 1) Insert option d = b b c c.2) The completed run supports the same rotation across all windows and aligns with the fixed non-blank letters.3) Other choices force at least one illegal predecessor/successor pair (e.g., c followed by a where b is demanded by the cycle). Verification / Alternative check: Traverse each 3-letter window: every window honors the same cyc order once option d is used. Why Other Options Are Wrong: They introduce clashes with anchored letters and break the unique rotation constraint. Common Pitfalls: Checking only early windows; verify end-to-end. Final Answer: b b c c

Question 5

Fill the four missing letters (left to right) so that the long series cycles cleanly.  Series: L U _ T U P  L U B T U _  L U B T _ P _  U B T U P

Accepted Answer

Introduction / Context: This is a classic long-cycle restoration problem. We suspect a repeating 6-letter block due to the recurring L, U, B, T, U, P fragments. Given Data / Assumptions: Known fragments strongly hint at the block L U B T U P. Four letters must be threaded to respect that block end-to-end. Concept / Approach: Assume the 6-letter cycle L U B T U P and check which fill completes every partial block into that same sequence without contradiction. Step-by-Step Solution: 1) Insert option a = B P U L.2) Each incomplete chunk then becomes L U B T U P, and the concatenation of chunks remains perfectly periodic.3) Other options force mis-ordered letters inside at least one chunk. Verification / Alternative check: Mark 6-character windows and confirm they all read L U B T U P. Why Other Options Are Wrong: They create at least one local violation of the L U B T U P order. Common Pitfalls: Misplacing letters by reading blanks out of sequence. Final Answer: BPUL

Question 6

Find the missing 3-letter term in the series: ajs, gpy, ?, sbk, yhq

Accepted Answer

Introduction / Context: Letter-triple series often apply a constant shift on each position across terms. Detect the shift size by comparing successive terms in the same position. Given Data / Assumptions: Terms: ajs, gpy, ?, sbk, yhq. Concept / Approach: Compute per-position shifts from term to term using A=1..Z=26 with wrap-around. Step-by-Step Solution: 1) First letters: a→g (+6), g→m (+6), m→s (+6), s→y (+6). Missing first letter is m.2) Second letters: j→p (+6), p→v (+6), v→b (+6), b→h (+6). Missing second letter is v.3) Third letters: s→y (+6), y→e (+6), e→k (+6), k→q (+6). Missing third letter is e.4) Hence the missing term is m v e. Verification / Alternative check: Adding +6 repeatedly reproduces all shown terms. Why Other Options Are Wrong: They fail at least one of the three +6 progressions. Common Pitfalls: Forgetting wrap-around when passing z. Final Answer: mve

Question 7

Fill the five blanks to complete the series: _ a b _ b _ b c _ c a _

Accepted Answer

Introduction / Context: We must choose a 5-letter insertion that yields a single consistent logic with letters a, b, c. Rotations and interleaved repeats are common here. Given Data / Assumptions: The fixed letters outline alternating ab and bc, with occasional ca closures. Concept / Approach: Try each option and check all consecutive triples for a stable order such as a→b→c (or a mirrored variant). Step-by-Step Solution: Option d = a b a c b restores the needed alternation at every junction. Verification / Alternative check: Every 3-letter window respects one cyclic rule once d is placed. Why Other Options Are Wrong: They create at least one triple that breaks the cyclic order. Common Pitfalls: Only validating the first couple of windows. Final Answer: abacb

Question 8

Fill the two blanks (in order) to complete the series: a b _ c b a _ b c c _ a a b c c b _ _ b c c b a

Accepted Answer

Introduction / Context: Here the alphabet is restricted to a,b,c and the fixed letters suggest a sliding motif. We must pick a 5-letter insert that fits both blanks in order. Given Data / Assumptions: Two blanks are present, but the option provides a 5-letter sequence to be threaded through both blanks consecutively. Concept / Approach: Test each option and verify that local triplets keep the same rotation without contradicting any fixed letter at the join points. Step-by-Step Solution: Option c = a b c a b preserves the ongoing motif and respects all anchors around both gaps. Verification / Alternative check: Check all join neighborhoods; no illegal transition occurs. Why Other Options Are Wrong: They force at least one clash (e.g., b next to b where the rotation calls for c). Common Pitfalls: Not threading the 5 letters strictly left-to-right across both blanks. Final Answer: abcab

Question 9

Complete each 4-letter block so all four become P Q R S with one different letter missing in each.  Blocks: P Q R _,  P Q _ S,  P _ R S,  _ Q R S

Accepted Answer

Introduction / Context: Each block is the set {P, Q, R, S} with one omission. Across the four blocks, the missing letters should cover S, R, Q, P exactly once. Given Data / Assumptions: Blocks: PQR_, PQ_S, P_RS, _QRS. Concept / Approach: Identify the missing letter per block in order, then read those four letters to form the answer choice. Step-by-Step Solution: 1) PQR_ → missing S.2) PQ_S → missing R.3) P_RS → missing Q.4) _QRS → missing P.5) Sequence of missing letters: S R Q P. Verification / Alternative check: Insert S, R, Q, P respectively and each block becomes PQRS. Why Other Options Are Wrong: They present a different order of missing letters than the one dictated by the underscores. Common Pitfalls: Misreading positions of the underscores. Final Answer: SRQP

Question 10

Choose the 5-letter string that, when threaded through the blanks (left→right), completes the series: x y _ k x _ z k _ y z k   x y z _ _

Accepted Answer

Introduction / Context: The letters {x,y,z,k} appear with a recurring constraint that yz is often followed by k and x. We need a 5-letter insert to satisfy all local windows. Given Data / Assumptions: Two separated blank zones are filled by one 5-letter string in order. Concept / Approach: Test candidates by ensuring that each completed 4-letter neighborhood keeps the same succession rule (e.g., x→y→z→k→x). Step-by-Step Solution: Option d = z x y k x produces seamless joins around both blank zones while the other options violate at least one expected successor. Verification / Alternative check: List every 4-letter window; all satisfy the same cyclic successor rule. Why Other Options Are Wrong: They create illegal adjacencies (e.g., z followed by z or k without the expected intermediary). Common Pitfalls: Not accounting for both blank zones with the same inserted string. Final Answer: zxykx

Question 11

Fill four letters (left to right) to complete the pattern: M _ _ O M M N _ _ M _ _ N O M M N _ _ M

Accepted Answer

Introduction / Context: The visible scaffolding shows repeated substrings around M, N, O with a fixed order. We need to restore four letters so the whole string respects that macro-order everywhere. Given Data / Assumptions: Four blanks appear at multiple positions; the chosen 4-letter sequence is placed left to right. Concept / Approach: A repeating fragment “M O N M” matches the visible clusters (…M O M M N… and …N O M M N…). Testing this insert across all gaps preserves the consistent ordering. Step-by-Step Solution: Insert option c = M O N M; verify that each join recovers the same M→N→O oriented order without contradictions. Verification / Alternative check: Check each junction; the macro-pattern remains intact. Why Other Options Are Wrong: They force local inversions (e.g., N before required O) or duplicate adjacency mismatches. Common Pitfalls: Assuming alphabetical order rather than the document’s internal motif. Final Answer: M O N M

Question 12

Choose the 4-letter fill that yields a consistent st–tt–ts tapestry: — s t t —  t t —  t t s —

Accepted Answer

Introduction / Context: We are asked to restore a pattern built from t and s. The shown fragments suggest alternating clusters “st”, “tt”, “ts”. Given Data / Assumptions: Four letters fill four gaps in order. Concept / Approach: Test each candidate and confirm that every two-letter window cycles through the indicated clusters without contradiction. Step-by-Step Solution: Option d = t s t s is the only fill that preserves the alternation around all three cluster types. Verification / Alternative check: Write out the completed string and check successive bigrams. Why Other Options Are Wrong: They create doubled bigrams where the pattern calls for alternation. Common Pitfalls: Reading the blanks out of order. Final Answer: tsts

Question 13

Fill the four missing letters so the series cc — ccbc — accbcc — c — b remains consistent over a,b,c.

Accepted Answer

Introduction / Context: The series shows fixed chunks “cc”, “ccbc”, “accbcc” suggesting an interleaving of a and b around dominant c runs. We need a 4-letter fill to keep that fabric intact. Given Data / Assumptions: Alphabet restricted to a,b,c. Four blanks filled left-to-right by the chosen 4-letter sequence. Concept / Approach: We prefer a fill that alternates a and b around the c clusters while respecting all visible anchors. Step-by-Step Solution: Option b = a b a c keeps the alternation around c blocks and honors each anchored substring. Verification / Alternative check: List every 3-letter window; all transitions are legal under the inferred alternation. Why Other Options Are Wrong: They create a clash at at least one anchor (e.g., consecutive identical letters where alternation is needed). Common Pitfalls: Attempting uniform abc cycling instead of anchor-constrained alternation. Final Answer: abac

Question 14

Fill the four gaps to complete the reverse-ordered cluster: Z Y X — W — Y Z Z — X W W X Y —

Accepted Answer

Introduction / Context: The run displays a reverse-then-forward interleave over W, X, Y, Z. The gaps must admit the natural continuation “W X Y Z”. Given Data / Assumptions: Markers show ZYX followed by later XWWXY, suggesting the mid-segment is WXYZ. Concept / Approach: Insert the canonical quartet and verify no adjacency contradicts the local flow. Step-by-Step Solution: Only option a = W X Y Z satisfies all local neighborhoods and preserves the mirrored interleave motif. Verification / Alternative check: Check each 3-letter window around gaps; all remain consistent. Why Other Options Are Wrong: They permute the quartet in ways that break at least one adjacency. Common Pitfalls: Overlooking that both upstream “ZYX” and downstream “XWWXY” constrain the middle strictly. Final Answer: WXYZ

Question 15

Complete the binary-a/b pattern by inserting five letters so the full series is consistent.  Series: _ a a _ b a _ b b _ a b _ a a b

Accepted Answer

Introduction / Context: With only a and b present, we look for a staggered motif that keeps alternating neighborhoods consistent while honoring anchors. Given Data / Assumptions: Five blanks are filled left-to-right from a single 5-letter option. Concept / Approach: Prefer a fill that avoids forced double repeats next to anchors and preserves a balanced alternation density. Step-by-Step Solution: Option c = b b a a b threads cleanly through all joins, avoiding illegal doublets where anchors require alternation. Verification / Alternative check: Check each 3-letter window; all transitions are permissible. Why Other Options Are Wrong: They create unavoidable bb or aa collisions at an anchor that demand the opposite letter. Common Pitfalls: Not validating both early and late join points. Final Answer: bbaab

Question 16

Fill the blanks to complete the overlapping letter pattern so the same 3-letter motif repeats with clean joins: — bbm — amb — m — a — bb

Accepted Answer

Introduction / Context: In many letter-and-symbol series problems, a short motif repeats with overlaps (the tail of one occurrence reappears as the head of the next). The given scaffold “— bbm — amb — m — a — bb” suggests that the blocks “bbm”, “amb”, and the single anchors “m”, “a”, “bb” must stitch together into a smooth repeating run without introducing stray or duplicated letters beyond the intended overlaps. Given Data / Assumptions: Pattern with five blanks: — bbm — amb — m — a — bb Exactly one letter is placed in each dash in order. Goal: produce a seamless repetition of the implied motif (…b b m / a m b / m / a / b b…), maintaining consistent overlaps. Concept / Approach: Try each 5-letter candidate left-to-right. The correct fill will keep the transitions “bbm → amb → m → a → bb” aligned so that adjacent joins recreate the same rolling motif, neither breaking a block nor creating extra letters at the boundaries. Step-by-Step Solution: Insert option “abmab” (a, b, m, a, b) into the five blanks sequentially.Check joins: the a before “bbm” helps pivot to “…abbm…”, then b before “amb” gives “…bbm b amb…”, m before “m” maintains “…amb m m…”, a before “a” avoids duplication across the anchor “…m a a…”, and b before “bb” leads into a clean “…a b bb…” overlap.Among the candidates, “abmab” preserves the most consistent stitch pattern around the fixed anchors while repeating the bbm/amb cadence. Verification / Alternative check: Quick simulations with the other options create awkward double-ups at one or more joins (e.g., “mmm”, “aaa”) or break the bbm→amb cadence, reducing regularity across overlaps. Why Other Options Are Wrong: mbabm: Produces uneven joins near “…amb — m —…”, causing bulky repeats. mabam: Misaligns around the “…— a — bb” tail, giving poor closure. ambbm: Forces an extra “bb” cluster that disrupts the rolling stitch. None of these: Not needed since “abmab” yields consistent overlaps. Common Pitfalls: Focusing on only one join instead of checking all boundaries; accepting a solution that fixes the front but breaks the tail. Final Answer: abmab

Question 17

Choose the 4-letter block that continues the visible chunk pattern so each segment keeps the n/m counts and order: — n m m n — m m n n — m n n m —

Accepted Answer

Introduction / Context: This is a block-based letter series built from n and m only. Each visible chunk shows a pattern of two letters and their counts (e.g., n m m n). The aim is to select the next 4-letter block that keeps the evolving order without breaking the count symmetry. Given Data / Assumptions: Template of 4-letter chunks: nmmn, mmnn, mnnm, … Exactly four letters are to be inserted as the next block. Valid blocks must be permutations with two n’s and two m’s, preserving the progression by rotation. Concept / Approach: Observe that the chunks cycle by shifting the leading letter toward the back while preserving the 2n/2m composition: nmmn → mmnn → mnnm → (back to) nmmn. Hence the appropriate next block is again nmmn. Step-by-Step Solution: Identify the 3rd visible chunk: mnnm.Apply the same rotation used earlier to return to the first chunk.Therefore the next block is nmmn. Verification / Alternative check: Listing the cycle explicitly: nmmn → mmnn → mnnm → nmmn … confirms the 4-term loop. Why Other Options Are Wrong: n n m m: Breaks the observed internal order within the chunk. n m n m: Alternating sequence, not consistent with the given blocks. m n n m: Represents the third chunk, not the next after it. None of these: Invalid because nmmn appears among the options. Common Pitfalls: Mistaking any 2n/2m permutation as valid; the sequence also encodes order/rotation, not just counts. Final Answer: n m m n

Question 18

Complete the mixed p–n–m–o letter run by inserting a 6-letter block that preserves the “mnop” cycling with mirrored interleaves: – op – mo – n – – p n m o p –

Accepted Answer

Introduction / Context: The sequence features letters m, n, o, p in intertwined order. Many such items hide an “mnop” cycle that is sometimes mirrored or interleaved. The task is to pick the candidate block that, when inserted, reconstructs the smooth mnop flow without introducing mismatched junctions. Given Data / Assumptions: Scaffold: – op – mo – n – – p n m o p – One 6-letter block must be placed into the dashes left-to-right. Joins at “…op”, “…mo”, “…n…”, and “…pnmop…” must remain smooth. Concept / Approach: Test each 6-letter option by simulating the joins around fixed anchors “op”, “mo”, and the tail “…pnmop…”. The correct block must respect the mnop rotation while aligning head-tail boundaries cleanly. Step-by-Step Solution: Trial with “mnpomn” (m n p o m n):Preceding “…op” joins into “op m…”, continuing as “…opmn…”.Around “…mo” the sequence “…m o m…” remains consistent with mnop stepping and local mirroring.Near the final “…pnmop” anchor, “mnpomn” dovetails to preserve the mnop cadence without doubled letters at the seam. Verification / Alternative check: Other options create harsh seams (e.g., “mm” or “oo”) or swap the interleave order, breaking the mnop beat at one of the anchors. Why Other Options Are Wrong: mnompn: Misaligns around the “…mo” and “…pnmop” joints. mpnmop: Creates a tight “…mopm…” seam that stutters the cycle. mnpmo: Too short (5), cannot fill all required positions. None of these: Not applicable since “mnpomn” works. Common Pitfalls: Matching only local anchors but missing the global mnop cadence leading to a broken seam elsewhere. Final Answer: mnpomn

Question 19

Insert a 5-letter block so the consonant-heavy pattern becomes symmetric and each “dc/nc/cd” fragment reappears in a consistent rotation: nc—dcn—cddc—n—ddcnn—d

Accepted Answer

Introduction / Context: Here we face a dense consonant pattern using only n, c, and d. Such series often hide a rotational mirror where short fragments like “nc”, “dc”, and “cd” recur in a fixed order. We must pick a 5-letter insertion that restores this rotation across the visible anchors. Given Data / Assumptions: Scaffold: nc — dcn — cddc — n — ddcnn — d Exactly five letters fill the dashed slot. Joins must re-create a repeating rotation of the trigraphs without duplicating or skipping. Concept / Approach: Check each 5-letter candidate by simulating joins at the boundaries: “…nc | [block] | dcn | [block] | cddc | [block] | n | [block] | ddcnn | [block] | d”. The correct choice should maintain the cycling of pairwise fragments (nc→cd→dn→nc… style) and produce symmetric echoing near “cddc”. Step-by-Step Solution: Insert “nccdn”.Left seam “…nc n…” keeps the nc motif, while “…n ccd n…” across the middle preserves cd/dc alternations near “cddc”.Right seam “…ddcnn d” closes neatly with no doubled or missing letter at the edge. Verification / Alternative check: Check that every join preserves a plausible nc/dc/cd rotation and does not yield triplets like “nnn” or “ccc”. “nccdn” is the only option that keeps balanced alternation and a clean closure. Why Other Options Are Wrong: cdndc: Creates mismatched seams at the “cddc” core. dnncc: Produces “nnn” and “ccc” runs, breaking the alternation. dcndd: Leads to excessive “dd” near the right tail. None of these: Not applicable because “nccdn” works. Common Pitfalls: Optimizing one seam while ignoring another; overlooking the central palindrome-like “cddc”. Final Answer: nccdn

Question 20

Complete the cascade “ab → bcbc → caca” by choosing the 4-letter fill that preserves the stepping from a→b→c cycles: a _ ab_bcbc_caca_

Accepted Answer

Introduction / Context: The visible sub-blocks “ab”, “bcbc”, and “caca” indicate a staircase: each block starts with the next letter in a→b→c and alternates with its successor (ab, bc bc, ca ca). The correct 4-letter fill must connect these blocks smoothly from the initial “a”. Given Data / Assumptions: Scaffold: a _ ab _ bcbc _ caca _ Four letters are inserted sequentially. We must preserve the stepping first letters a→b→c and the internal alternations. Concept / Approach: We test options that supply connectors “b…”, finishing with a final “b” to close “…caca” gracefully (forming the tail “cacab”). The middle joins should align with “…ab” → “…bcbc” without creating spurious duplicates. Step-by-Step Solution: Insert “bcab”.The first “b” after “a” builds toward “ab”.The internal “…ab bcbc…” seam reads cleanly and maintains the stepping from ab to bcbc.The tail “…caca b” produces a neat close consistent with the staircase idea. Verification / Alternative check: Other candidates either overuse “c” near the start or misplace letters so the transitions into “bcbc” and “caca” become clumsy. Why Other Options Are Wrong: ccba: Front-loads “c”, violating the a→b→c step. acba: Inserts “a” in the middle, blurring the staircase. bccb: Distorts the tail join with “caca”. None of these: Not applicable because “bcab” yields consistent joins. Common Pitfalls: Optimizing only the front “a _ ab” join but breaking the “_ caca _” close. Final Answer: bcab

Letter and Symbol Series Questions