Series (Number, Letter & Alphanumeric)
In short: Spot the rule in number, letter and alphanumeric sequences — and the odd/missing term.
Reasoning — Constable › Verbal & Analytical Reasoning · 34 min read · Free to read · English & Telugu
Bilingual study notes for the Andhra Pradesh Police SI & Constable exams (SLPRB AP) by AP Police Hub, a product of Anvil Academy.
Spot the hidden rule and continue the pattern — the everyday number-series families (gaps, powers, mixed operators, primes, Fibonacci, wrong & missing terms), the letter-series ladder, and simple alphanumeric sequences, each worked step by step.
1. Introduction & why it matters
A series is a sequence of numbers, letters, or a mix of both, arranged by a fixed hidden rule. Your job is to discover that rule from the terms given and then either fill the blank (the next or a missing term) or spot the odd term that breaks the rule. In the APSLPRB Constable Prelims the Reasoning section always carries a few series questions, and they are pure logic: once you know the standard families, most fall in under a minute.
Series questions are fast only if the building blocks are already in your head. Memorise the squares and cubes, the first primes, and — for letter/alphanumeric series — the alphabet-position table. Then recognising a pattern becomes instant.
Squares 1²–15²| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
|---|
| 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 | 121 | 144 | 169 | 196 | 225 |
Cubes 1³–10³| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|
| 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 |
The alphabet-position table (for letter series)
For every letter and alphanumeric series you must place a letter instantly. Learn the forward position (A = 1 … Z = 26) and lean on the EJOTY milestones — E = 5, J = 10, O = 15, T = 20, Y = 25 — then step a little forward or back.
Forward position (A = 1 … M = 13)| A | B | C | D | E | F | G | H | I | J | K | L | M |
|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
Forward position (N = 14 … Z = 26)| N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
|---|
| 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 |
3. Number series I — one operator
The simplest series apply the same single operation to jump from each term to the next: add a constant, subtract a constant, multiply by a constant, or divide by a constant. Always test these four first — most easy questions are one of them.
Model 1 — Constant difference (add / subtract)
Model 1a — add a constant
Question: Find the next term: 8, 13, 18, 23, 28, ?
Approach: Before guessing anything, the safest first move on any series is to write the gap between each pair of neighbours. If that gap stays the same all the way along, the rule is simply 'add a constant', and the next term is just one more gap beyond the last.
13-8=5,\ 18-13=5,\ 23-18=5,\ 28-23=5 — Take the difference of every neighbouring pair. Each one comes out to 5 exactly, so the series climbs by the same step of 5 every time — a constant-difference (arithmetic) series:28+5=33 — Since the step never changes, extend it once more: add 5 to the last given term 28 to reach the term that follows:
Answer: The gap is a steady +5, so the term after 28 is 33.
Model 1b — subtract a constant
Question: Find the next term: 50, 44, 38, 32, 26, ?
Approach: The numbers here are getting smaller, and they slide down gently rather than dropping suddenly. A slow, even fall is the signature of a constant subtraction, so measure how much is lost at each step.
50-44=6,\ 44-38=6,\ 38-32=6,\ 32-26=6 — Subtract each term from the one before it. Every gap is exactly 6, so the same amount is taken away at each step:26-6=20 — Carry the same fall one step further — take 6 away from the last term 26:
Answer: The series falls by a fixed 6 each time, so the term after 26 is 20.
Model 2 — Constant ratio (multiply / divide)
Model 2a — multiply by a constant
Question: Find the next term: 2, 10, 50, 250, ?
Approach: Here the numbers do not creep up — they leap: 2 to 10 to 50. When a series grows this fast, addition cannot explain it, so instead of a difference we look at the ratio, dividing each term by the one before to see if it is being multiplied by a fixed number.
10\div2=5,\ 50\div10=5,\ 250\div50=5 — Divide each term by its predecessor. The answer is 5 every time, so each term is exactly five times the one before — a constant-ratio (geometric) series:250\times5=1250 — Apply the same ratio once more — multiply the last term 250 by 5:
Answer: Each term is 5 times the last, so the term after 250 is 1250.
Model 2b — divide by a constant
Question: Find the next term: 2187, 729, 243, 81, ?
Approach: The terms tumble down steeply — 2187 to 729 to 243 — far too fast for plain subtraction. A sudden, steep fall is the mirror image of a multiplying series, so we test division: divide each term by the next to find the constant factor being removed.
2187\div3=729,\ 729\div3=243,\ 243\div3=81 — Divide each term by the one after it. Every result is 3, so at each step the number is cut to a third of its value:81\div3=27 — Continue the same rule — divide the last term 81 by 3:
Answer: Each term is one-third of the last, so the term after 81 is 27.
4. Number series II — changing gaps
When the difference is not constant, look at how the differences themselves change. If the gaps grow (or shrink) by a fixed amount, the series is a changing-gap series. Write the gaps under the terms and the rule almost always shows itself.
Model 3 — Growing gap (differences increase)
Model 3a — increasing difference
Question: Find the next term: 3, 5, 9, 15, 23, ?
Approach: When you write the gaps here they are 2, 4, 6, 8 — not a single fixed number, so this is not a plain add-a-constant series. But the gaps themselves change in a tidy way, and that second pattern is the real rule; find how the gaps grow and the next term follows.
+2,\ +4,\ +6,\ +8 — List the differences between neighbours. They are 2, 4, 6, 8, each one bigger than the last by 2 — the gaps march up in a steady step:23+10=33 — So the gap that comes next must be 8+2=10. Add this new gap to the last term 23:
Answer: The gaps grow 2,4,6,8,10, so the term after 23 is 23+10=33.
Model 3b — decreasing difference
Question: Find the next term: 2, 15, 25, 32, 36, ?
Approach: The series still rises, but by less and less each time. That tells you the gaps are shrinking rather than growing, so look at the differences and check whether they fall by a fixed amount — if they do, the next gap is easy to predict.
+13,\ +10,\ +7,\ +4 — Write out the gaps: 13, 10, 7, 4. Each is 3 smaller than the one before, so the differences are stepping down evenly:36+1=37 — Continuing the fall, the next gap is 4-3=1. Add that final small gap to the last term 36:
Answer: The gaps shrink 13,10,7,4,1, so the term after 36 is 36+1=37.
Shortcut: Whenever the gaps march in an even step (here −3 each time), just carry that same step one line further to get the next gap.
5. Number series III — squares & cubes
A sudden, clean jump that is not a simple ratio almost always hides a square or cube pattern, usually with a small ±1 (or ±2) tweak. Compare each term to the nearest perfect square or cube — the offset is the rule.
Model 4 — Squares with a ±k tweak
Model 4
Question: Find the next term: 3, 6, 11, 18, 27, ?
Approach: The jump from 3 to 6 to 11 is clean but does not fit a single ratio, and a jump like this almost always hides a power. Compare each term with the nearest perfect square: if every term sits the same small distance above a square, the rule is n^2 plus a fixed offset.
1^2+2=3,\ 2^2+2=6,\ 3^2+2=11,\ 4^2+2=18,\ 5^2+2=27 — Line the terms up against the squares 1, 4, 9, 16, 25. Every term is exactly 2 more than the matching square, so the rule is n^2+2 with n = 1, 2, 3, \dots:6^2+2=36+2=38 — The fifth term used 5^2, so the sixth uses the next square 6^2 = 36, then add the same 2:
Answer: Every term is a square plus 2, so the next is 6^2+2=38.
Shortcut: Whenever a term is a little more than a square (3, 6, 11, 18, 27 → 1, 4, 9, 16, 25), the rule is n^2+k; read k from any one term.
Model 5 — Cubes with a ±k tweak
Model 5
Question: Find the next term: 3, 10, 29, 66, 127, ?
Approach: This series shoots up much faster than a squares series would — 3, 10, 29, 66, 127 — which is the tell-tale sign of cubes. Set each term beside the cubes 1, 8, 27, 64, 125 and read off the small constant that separates them.
1^3+2=3,\ 2^3+2=10,\ 3^3+2=29,\ 4^3+2=66,\ 5^3+2=127 — Each term is exactly 2 above the matching cube, so the rule is n^3+2 with n = 1, 2, 3, \dots:6^3+2=216+2=218 — The fifth term used 5^3, so the next uses the cube 6^3 = 216, and we add the same 2:
Answer: Every term is a cube plus 2, so the next is 6^3+2=218.
Shortcut: Terms near 1, 8, 27, 64, 125 (the cubes) signal a cube rule — read the small offset from any one term.
6. Number series IV — mixed & compound
Compound series use two operators at each step — most often 'multiply then add a constant' (\times k + c). Others alternate the operator each step. When a plain ratio almost fits but leaves a small remainder, look for a \times k + c rule.
Model 6 — Multiply then add (× k + c)
Model 6
Question: Find the next term: 4, 11, 25, 53, ?
Approach: If you divide 11 by 4 you get a little over 2, and the same near-doubling shows all along — but a plain \times 2 always leaves a small leftover. When a ratio almost fits but overshoots by a steady amount, the rule is usually 'multiply by k, then add a constant c'; here we test \times 2 + 3.
4\times2+3=11,\ 11\times2+3=25,\ 25\times2+3=53 — Try doubling each term and adding 3. It reproduces every following term exactly, so the compound rule \times 2 + 3 is confirmed:53\times2+3=109 — Use the same two-part rule on the last term 53 — double it and add 3:
Answer: Applying \times 2 + 3 to 53 gives the next term 109.
Shortcut: For a \times k + c series the gaps between terms themselves grow by a factor of k (here 7, 14, 28 each double). Spot that doubling of the gaps and you have confirmed k=2 before even hunting for c.
Model 7 — Alternating operations (× then ÷)
Model 7
Question: Find the next term: 6, 24, 12, 48, 24, ?
Approach: Notice the terms do not move in one direction — they jump up to 24, drop to 12, jump to 48, drop to 24. A saw-tooth like this means two different operations are taking turns, one at the odd steps and another at the even steps, so read each step on its own.
6\times4=24,\ 24\div2=12,\ 12\times4=48,\ 48\div2=24 — Read the operations in turn: an up-step is \times 4 and the following down-step is \div 2, and this pair repeats:24\times4=96 — The last operation used was a \div 2, so the turn now comes back to \times 4 on the last term 24:
Answer: The operators alternate \times 4, \div 2, and a \times 4 is due next, so the term after 24 is 96.
Shortcut: When a series zig-zags up and down, split it into the 'up' step and the 'down' step and read each separately.
7. Number series V — special sequences
A handful of series follow a named idea rather than a plain arithmetic rule: prime numbers, Fibonacci-type (add the two before), two series interleaved into one, or a rule that acts on the digits of each term. Recognise the signature and the answer follows at once.
Model 8 — Prime-number series
Model 8a — the primes themselves
Question: Find the next term: 5, 7, 11, 13, 17, ?
Approach: The gaps here are 2, 4, 2, 4 — they do not settle to one value and they do not grow smoothly, so no arithmetic rule fits. Whenever the differences refuse to make sense, step back and ask whether the numbers themselves are a famous list; here they are the prime numbers in order.
- Each term 5, 7, 11, 13, 17 has exactly two factors, so the series is simply the prime numbers listed one after another.
\text{next prime}=19 — So the answer is the very next prime after 17, which is 19 (note 18 is skipped because it is not prime):
Answer: The terms are consecutive primes, so the one after 17 is 19.
Model 8b — add the next prime
Question: Find the next term: 4, 6, 9, 14, 21, ?
Approach: The terms themselves are not primes, but the moment you write the gaps you notice something familiar: 2, 3, 5, 7. The differences are the prime numbers in order, so the rule is 'add the next prime each time'.
+2,\ +3,\ +5,\ +7 — Confirm the gaps between neighbours are 2, 3, 5, 7 — the first four primes, each added in turn:21+11=32 — The prime after 7 is 11, so add 11 to the last term 21:
Answer: The gaps are successive primes, and the next is +11, so the term after 21 is 32.
Model 9 — Fibonacci-type (add the two before)
Model 9
Question: Find the next term: 2, 5, 7, 12, 19, ?
Approach: The gaps here jump around and no single operator ties one term to the next, but a Fibonacci-type series links each term to the two before it rather than just one. When you spot that a term looks like the sum of the previous pair, test that idea across the whole series.
2+5=7,\ 5+7=12,\ 7+12=19 — Add each pair of neighbours and check it makes the term after them: 2+5=7, 5+7=12, 7+12=19 — every one lands on the next term, so the rule holds:12+19=31 — So the next term is the sum of the last two given terms, 12 and 19:
Answer: Each term is the sum of the previous two, so after 12 and 19 comes 31.
Model 10 — Two interleaved series
The odd-placed terms form one series and the even-placed terms another; they are woven together. Split them apart and solve each.
Model 10
Question: Find the next term: 4, 30, 8, 25, 12, 20, 16, ?
Approach: One term goes up, the next goes down, then up again — as if two different series have been shuffled together like a zip. The cure is to unzip them: the terms in the odd positions form one series and those in the even positions another, and each obeys its own simple rule.
4,\ 8,\ 12,\ 16\quad(+4) — Pick out the odd-position terms — the 1st, 3rd, 5th, 7th. They are 4, 8, 12, 16, a plain series rising by 4:30,\ 25,\ 20,\ ?\quad(-5) — Now the even-position terms — the 2nd, 4th, 6th, and the missing 8th. They are 30, 25, 20, ?, falling by 5 each time:20-5=15 — The blank we want is the 8th term, which sits in the even series, so continue that falling line: 20-5:
Answer: The 8th term belongs to the even series 30,25,20,\dots, so it is 20-5=15.
Shortcut: Two interleaved series usually reveal themselves by size: if the terms alternate small–large or up–down, take every second term as one line and solve the two lines apart.
Model 11 — Digit rotation
Model 11
Question: Find the next term: 5678, 8567, 7856, 6785, ?
Approach: Every term here uses the very same four digits 5, 6, 7, 8 — only their order changes. When the digits stay the same but shuffle, the rule is not arithmetic at all; it is a rearrangement, so compare where each digit sits from one term to the next.
567\underline{8}\to\underline{8}567 — Compare 5678 with 8567: the last digit 8 has jumped to the front and everything else has slid one place right. So each step takes the last digit round to the front:678\underline{5}\to\underline{5}678 — Apply the same move to the last term 6785: its final digit is 5, so bring that 5 to the front:
Answer: Moving the last digit of 6785 to the front gives 5678.
8. Wrong term & missing middle term
Two twists appear often. In a wrong-term question the whole series obeys one rule except one intruder — find it. In a missing-middle question the blank sits inside the series, not at the end — you must read the rule from both sides of the gap.
Model 12 — Find the wrong term
Model 12a — wrong term in an add series
Question: One term is wrong. Find it: 3, 8, 13, 18, 24, 28.
Approach: In a wrong-term question all but one of the numbers obey a single rule, so do not try to force a pattern onto every term at once. Instead find the rule that the majority of the terms follow, then the one term that refuses to fit is your answer.
3\;(+5)\;8\;(+5)\;13\;(+5)\;18\;(+5)\;23\;(+5)\;28 — Start from the left and read the gaps: 3, 8, 13, 18 all step up by 5, so the intended rule is 'add 5', which would continue 23, 28:18+5=23\neq24 — The fifth term ought to be 18+5=23, yet the series prints 24 there — that single mismatch is the intruder (and the final 28 fits the true rule again):
Answer: The series should add 5 each time, so the wrong term is 24 — it ought to be 23.
Model 12b — wrong term in a × + series
Question: One term is wrong. Find it: 4, 9, 19, 39, 80, 159.
Approach: Each term is a little more than double the one before, and a near-doubling that leaves a small leftover points to a \times 2 + c rule. Fix the rule from the run of terms that agree, then hunt for the single term that breaks it.
39\times2+1=79 — Test 'double and add 1': 4\to9\to19\to39 all obey it, and continuing gives 39\times2+1=79, then 79\times2+1=159 — so \times 2 + 1 is the true rule:79\neq80 — The fifth term should therefore be 79, but the series shows 80; since the final 159 agrees with 79\times2+1, it is the 80 that is out of place:
Answer: Under the \times 2 + 1 rule the fifth term must be 79, so the wrong term is 80.
Shortcut: For a wrong-term question, trust the LONGEST run that obeys one rule; the term that breaks it is the intruder.
Model 13 — Find the missing middle term
Model 13a — missing middle, × 3
Question: Find the missing term: 2, 6, ?, 54, 162.
Approach: The blank sits inside the series, not at the end, but the method is the same: read the rule from the terms you can see on one side of the gap, fill the blank, and then confirm it against the terms on the other side.
2\times3=6 — The step from 2 to 6 is a multiplication by 3, so read the rule as \times 3 from the left of the gap:6\times3=18 — Apply that rule to the term just before the blank — multiply 6 by 3:18\times3=54\ \checkmark — Check that this value flows on into the terms after the gap: 18\times3=54 and 54\times3=162, both of which are given, so 18 is right:
Answer: The rule is \times 3, so the missing middle term is 6\times3=18.
Model 13b — missing middle read from both sides
Question: Find the missing term: 2, 6, 12, ?, 30, 42.
Approach: Here the gaps are not constant, so first work out how they grow. Once the pattern of gaps is clear you can reach the blank from the left; because a middle term can always be checked from both directions, close in from the right as well and make sure the two answers agree.
+4,\ +6,\ +8,\ +10,\ +12 — Write the gaps between the visible terms: they are the even numbers 4, 6, 8, 10, 12, each 2 more than the last:12+8=20 — The gap into the blank (the third gap) is +8, so add it to the term before, 12:30-10=20\ \checkmark — Confirm from the other side: the gap leaving the blank is +10, so 30-10 must give the blank too — and it gives 20, matching exactly:
Answer: Both directions meet at 20, so the missing term is 20.
Shortcut: For a missing middle, work the rule inward from BOTH ends — the two directions must meet at the same value.
9. Letter series
A letter series moves through the alphabet by a rule. Convert each letter to its position (Section 2), read the pattern in the positions exactly as you would for numbers, then convert back. The common forms are a fixed skip, a growing skip, a reverse skip, and clusters of two letters moving together.
Model 14 — Constant skip (+n)
Model 14
Question: Find the next letter: B, G, L, Q, ?
Approach: A letter series is just a number series in disguise. Turn each letter into its alphabet position, read the pattern in those numbers exactly as you would for a number series, then convert the answer back to a letter.
2,\ 7,\ 12,\ 17\quad(+5) — Replace each letter by its position: B=2, G=7, L=12, Q=17. These climb by a steady 5, a plain constant-skip series:17+5=22=V — Add one more 5 to reach position 22, and the 22nd letter is V:
Answer: The positions rise by 5, so after Q (17) comes position 22 = V.
Model 15 — Growing skip (1, 2, 3, …)
Model 15
Question: Find the next letter: A, C, F, J, ?
Approach: Convert the letters to positions again, but this time do not expect a fixed skip — check whether the skip itself grows. When the jumps increase by one each step, the next jump is simply one larger.
+2,\ +3,\ +4 — The positions are 1, 3, 6, 10, and the jumps between them are +2, +3, +4 — each skip is one bigger than the last:10+5=15=O — So the next skip is +5, taking 10 to position 15, and the 15th letter is O:
Answer: The skip grows 2,3,4,5, so after position 10 comes 15 = O.
Model 16 — Reverse skip (−n)
Model 16
Question: Find the next letter: Y, U, Q, M, ?
Approach: These letters move backwards through the alphabet, so once again convert to positions and expect the numbers to fall. A fixed downward step is just a subtraction series in position-form.
25,\ 21,\ 17,\ 13\quad(-4) — In positions the letters are 25, 21, 17, 13, falling by a fixed 4 each time:13-4=9=I — Subtract one more 4 from position 13 to land on position 9, and the 9th letter is I:
Answer: The positions fall by 4, so after M (13) comes position 9 = I.
Model 17 — Two-letter clusters (two lines at once)
Each cluster has two letters; the first letters form one series and the second letters another. Track the two lines separately.
Model 17
Question: Find the next cluster: AZ, CX, EV, GT, ?
Approach: Each item is a pair of letters, and the two positions inside a pair have nothing to do with each other. Treat the first letters as one series and the second letters as a completely separate series, solve each in positions, then put the two answers back together.
1,\ 3,\ 5,\ 7\to9=I — The first letters are A, C, E, G — positions 1, 3, 5, 7 climbing by 2, so the next first letter is position 9 = I:26,\ 24,\ 22,\ 20\to18=R — The second letters are Z, X, V, T — positions 26, 24, 22, 20 falling by 2, so the next second letter is position 18 = R:
Answer: Joining the two lines, the next cluster is I with R, i.e. IR.
Shortcut: In a two-letter cluster, one letter usually climbs and the other usually falls — read them as two independent lines.
10. Alphanumeric series
An alphanumeric series mixes letters and numbers (sometimes symbols too). Two flavours appear. First, paired progressions — each term is a letter tied to a number, and the letter and the number each follow their own rule. Second, a long mixed string on which you answer position and counting questions.
Model 18 — Letter + number, same rule
Model 18
Question: Find the next term: B4, E7, H10, K13, ?
Approach: Each term carries a letter and a number bolted together, and the safe way to handle such a pair is to split it: follow the letters as one series and the numbers as another, then rejoin them at the end.
2,\ 5,\ 8,\ 11\to14=N — The letters B, E, H, K are positions 2, 5, 8, 11, rising by 3, so the next letter is position 14 = N:4,\ 7,\ 10,\ 13\to16 — The numbers 4, 7, 10, 13 also rise by 3, so the next number is 16; here the letter and number happen to share the same step:
Answer: Pairing the next letter with the next number gives N16.
Model 19 — Letter + number, different rules
Model 19
Question: Find the next term: 3A, 6C, 12E, 24G, ?
Approach: Again split the number from the letter, but this time do not assume they share a rule — the number part and the letter part each follow their own, so work out the two independently.
3,\ 6,\ 12,\ 24\to48 — The numbers 3, 6, 12, 24 double at every step (a \times 2 rule), so the next number is 24\times 2 = 48:1,\ 3,\ 5,\ 7\to9=I — The letters A, C, E, G are positions 1, 3, 5, 7 rising by 2 (a completely different rule), so the next letter is position 9 = I:
Answer: The number doubles to 48 and the letter advances to I, giving 48I.
Model 20 — Position & counting on a mixed string
Here you are given one long sequence of letters, numbers and symbols and asked position or counting questions about it. Number the positions from the left before you start. Use this sequence for the next three models:
Model 20a — nth from the right
Question: In the sequence above, which element is 5th from the right?
Approach: Rather than counting backwards from the right — where it is easy to slip — convert the position into a count from the left, which you have already numbered. An item that is n-th from the right is the (\text{total} - n + 1)-th from the left.
16-5+1=12 — The string has 16 elements, so the 5th from the right is at 16 - 5 + 1 = 12 from the left:- Count 12 places from the left in R\,3\,K\,\%\,D\,7\,M\,@\,B\,5\,W\,F\dots and you land on F:
Answer: The 5th from the right is the 12th from the left, which is F.
Shortcut: Left rank + right rank = total + 1. So the 5th from the right is the (16 − 5 + 1) = 12th from the left.
Model 20b — count letters followed by a number
Question: In the sequence above, how many letters are immediately followed by a number?
Approach: The phrase 'immediately followed by' means you only ever look at the single element sitting right after each letter — never one further along. So walk through the letters one by one and glance at the very next slot each time.
R,\ D,\ B,\ F — Going through the string, four letters have a digit right after them: R is followed by 3, D by 7, B by 5, and F by 8:- The other letters fail the test — K is followed by \%, M by @, W by F (a letter, not a number) and N by \# — so none of these count:
4 — Counting only the letters that passed gives the total:
Answer: Exactly 4 letters (R, D, B, F) are immediately followed by a number.
Model 20c — dropping elements
Question: If all the numbers are dropped from the sequence, which element is now 7th from the left?
Approach: Once elements are removed the old position numbers no longer apply, so the reliable move is to physically rewrite the shortened string first and only then count. Trying to count on the original string is where mistakes creep in.
- Strike out the digits 3, 7, 5, 8, 2 and write down what is left, keeping the order: R\ K\ \%\ D\ M\ @\ B\ W\ F\ N\ \#:
1{:}R,\ 2{:}K,\ 3{:}\%,\ 4{:}D,\ 5{:}M,\ 6{:}@,\ 7{:}B — Now count seven places from the left on this new, shorter string:
Answer: After the numbers are dropped, the 7th element from the left is B.
Shortcut: After dropping items, always rewrite the shortened string first — never count on the old positions.
11. Shortcuts, traps & tips
12. At a glance
Every family, its tell-tale clue and its method| Family | Tell-tale clue | Method |
|---|
| Constant difference | Same gap throughout | Add or subtract the gap |
| Constant ratio | Sudden even jump | Multiply or divide by the ratio |
| Changing gap | Gaps grow/shrink steadily | Extend the pattern of gaps |
| Squares / cubes ±k | Terms near 1,4,9… or 1,8,27… | Match to n^2 / n^3; read the offset |
| × k + c (compound) | Ratio close but not exact | Test multiply-then-add/subtract |
| Alternating operations | Terms zig-zag up and down | Read the up-step and down-step apart |
| Prime series | 2,3,5,7,11… or prime gaps | Use the next prime |
| Fibonacci-type | Term = sum of two before | Add the last two terms |
| Interleaved | Two patterns woven together | Split odd & even places |
| Digit-based | Next term built from digits | Rotate or work on the digits |
| Wrong / missing term | One odd term or an inner blank | Fix the rule, then find the mismatch |
| Letter series | Letters shift along the alphabet | Use positions; mind the Z→A wrap |
| Alphanumeric | Letters + numbers together | Track letter & number rules apart |