Series (Number, Letter and Mixed)

What the exam asks. Five or six numbers with small steps and a question mark at the end (sometimes in the middle). This is the commonest opening item of a series set, so make it your first tool.

Why it works. When numbers change slowly, each term was made from the one before by adding something. That "something" is the difference. So the plan is always the same: find the difference, see how it behaves, and extend it by one step.

If the jumps between terms are small compared with the numbers themselves, the pattern is almost always addition or subtraction. Your tool is the difference row: write each term minus the term before it, just under the series.

• All differences equal. The series adds (or subtracts) the same number every time. 7, 15, 23, 31 adds 8.
• Differences grow (or shrink) by a fixed amount. 4, 5, 7, 10, 14 has differences 1, 2, 3, 4, which go up by 1 each time. So the next difference is 5 and the next term is 14 + 5 = 19.
• Differences are not even regular. Write the differences of the differences (the second row). If that row is constant, the pattern is found. If not, go one row deeper.

[[fig:0]]

Falling series follow the same method, but carry the minus sign. In 60, 52, 45, 39 the differences are −8, −7, −6, so the next difference is −5 and the next term is 34. A common slip is to switch the sign and add.

When the second row is constant, work back up: extend the second row by one step, add it to the last difference, then add that difference to the last term.

Sometimes you need three rows. In 3, 4, 9, 24, 55 the rows are 1, 5, 15, 31 then 4, 10, 16 then 6, 6. The third row is constant, so the second row continues 22, the first row continues 31 + 22 = 53, and the next term is 55 + 53 = 108.

Write the difference rows. Stop only when a row is constant or you recognise it as a known list.

What the exam asks. The terms roughly double or triple at every step (3, 8, 18, 38 or 5, 14, 41, 122); or they sit in a plain ratio (2, 4, 8, 16 or 729, 243, 81, 27); or the multiplier itself keeps growing (28, 12, 10, 13, 24). You are asked for the next term.

Why it works. Each term is made from the one before by one multiplication and one fixed add-on. Because of that, the differences are multiplied by the same k at every step. That single fact lets you read k straight off the difference row (Trick 1 below) instead of guessing it.

When the terms jump by about double or more, subtraction rows will not help. Each term has been built from the one before by multiplying by k and then adding or subtracting c.

Take 3, 8, 18, 38, 78. The ratio of the large terms is close to 2, so try k = 2. Then 3 × 2 = 6 and we need 8, so c = +2. Check the rest: 8 × 2 + 2 = 18, 18 × 2 + 2 = 38, 38 × 2 + 2 = 78. The pattern fits every pair, so the next term is 78 × 2 + 2 = 158.

[[fig:0]]

How to find k and c quickly.
1. Divide the two largest terms to estimate k (78 ÷ 38 is about 2).
2. Use the first pair to get c: first term × k, then see what must be added or subtracted.
3. Check the pattern on every pair.

Variants you will meet:

• × k + c or × k − c: 5, 14, 41, 122 uses × 3 − 1.
• Halving with a fixed add-on (÷ 2 + c): 98, 50, 26, 14, 8 uses ÷ 2 + 1.
• × 1.5: 32, 48, 72, 108, 162. Go step by step; a slip on the last step is the usual mistake.
• Alternating × and ÷ in a row, such as × 2, ÷ 4, × 2, ÷ 4.

A growing multiplier. In 28, 12, 10, 13, 24, 58 the multipliers go ½, 1, 1½, 2, 2½ and the same − 2 follows each of them: 28 × ½ − 2 = 12, 12 × 1 − 2 = 10, 10 × 1½ − 2 = 13, 13 × 2 − 2 = 24, 24 × 2½ − 2 = 58. The clue is that the difference row (−16, −2, 3, 11) has no constant ratio.

A pattern is a pattern only if it fits every pair. "× 2 + 2" is not the same as "double".

If the ratio keeps changing (2.0, 2.2, 2.4…), you have a different pattern. Return to the differences.

What the exam asks. A series that goes up and down, or whose differences zig-zag: 80, 95, 75, 90, 70, ? or 391, 349, 376, 334, 361, 319, ?. Sometimes the blank is in the middle or there are two blanks. Railway papers set this type often.

Why it works. The paper-setter has woven two simple series together, one on the odd places and one on the even places. Each line alone is easy; the difficulty is only that they are mixed. Your whole job is to un-mix them.

Sometimes two independent series are woven together, one on the odd places (1st, 3rd, 5th …) and one on the even places (2nd, 4th, 6th …). The terms then jump up and down, or the series refuses to fit any single difference pattern.

Warning signs:

• The series rises and falls: 60, 5, 55, 10, 50 …
• The differences zig-zag between two kinds of values.
• No single pattern fits the whole series, but every second term looks regular.

Method.
1. Number the places 1, 2, 3, 4, … on paper.
2. Write the odd-place terms in one line and the even-place terms in another.
3. Solve each line separately.
4. Find which line the missing term lies on, and continue only that line.

[[fig:0]]

Take 4, 11, 6, 22, 8, 33, ?. Odd places: 4, 6, 8, so +2 each. Even places: 11, 22, 33, so +11 each. The 7th place is odd, so the answer is 8 + 2 = 10, not the even-line value 44.

Two question marks. In some questions two places are blank, say the 1st and the last. Solve the odd line and the even line independently; the two blanks usually come from different lines.

Different patterns in the two lines are allowed. One line may add, the other may square or double: 7, 4, 11, 9, 15, 16, 19 has +4 on the odd places and 2², 3², 4² on the even places, so the 8th term is 5² = 25.

Number the places before you choose the line. Most errors are position errors.

If the split gives no pattern on either line, return to a full-series approach (differences or a cycle).

What the exam asks. Two forms. (1) The terms themselves are perfect squares or cubes in a row, such as 1, 4, 9, 16, ? or 144, 121, 100, 81, ?. This is the commonest form in recent papers. (2) The terms are ordinary numbers, but their differences form a famous list, such as 1, 8, 27, 64 or 2, 3, 5, 7.

Form (1): name every term. Squares to 30 and cubes to 15 must be on your tongue. 144, 121, 100, 81 are 12², 11², 10², 9², so the next term is 8² = 64. The roots may rise, fall or skip (4, 16, 36, 64 are 2², 4², 6², 8²). The differences of a row of squares are the odd numbers 3, 5, 7, 9 …, because (n + 1)² − n² = 2n + 1, so the difference method also works, only more slowly.

Form (2): the differences are a famous list. Sometimes the differences are not equal and not an AP either, but they are a list you already know. Learn the lists below and the whole series falls open.

Take 3, 4, 8, 17, 33. The differences are 1, 4, 9, 16. These are 1², 2², 3², 4², so the next difference is 5² = 25 and the next term is 33 + 25 = 58.

[[fig:0]]

The differences can also be falling lists. In 499, 498, 490, 463, 399 the differences are −1, −8, −27, −64: cubes, subtracted. The next is −125, giving 274.

An alternating difference (+11, +9, +11, +9) is easy to miss. The series 20, 31, 40, 51, 60 has differences 11, 9, 11, 9. Split the steps into odd steps (all 11) and even steps (all 9).

Before you trust a special list, confirm it on at least three differences.

Not every irregular-looking row is special. The differences of the primes themselves (1, 2, 2, 4, 2…) look random, so do not hunt for a difference pattern there; recognise the primes as the terms.

What the exam asks. A series that rises in jerks: a small step, then a big one, then a small one again. Recent papers give 152, 140, 134, 122, 116, 104, ? (−12, −6, −12 …), 118, 149, 211, 242, 304, 335, ? (+31, +62 …), 26, 34, 45, 53, 64, 72, ? (+8, +11 …), and 8, 24, 12, 36, 18, 54, ? (× 3, ÷ 2 …).

Why it works. Two (sometimes three) operations are used in turn, in a fixed order. The odd-numbered steps share one operation and the even-numbered steps share the other. Once you have named the two, you only need to know where you are in the cycle.

Here the operation is not the same at every step. Two (or three) operations repeat in a fixed order. In 5, 11, 22, 28, 56 the steps are +6, ×2, +6, ×2. The odd steps are all "+6" and the even steps all "×2".

[[fig:0]]

How to spot it.

• The differences are irregular, but the series rises and falls in jerks: a small step, then a big one, then a small one.
• A single pattern (× k ± c) fails on the third pair.

How to solve it.
1. Look at the first step and guess its operation (add? multiply?).
2. Look at the second step and guess a different operation.
3. Test the guesses on step 3 and step 4. If step 3 repeats step 1's operation, you have a two-operation cycle.
4. Write the operation above each arrow, then continue the cycle. Take care of where you are in the cycle: after "+6, ×2, +6, ×2" the next operation is +6, not ×2.

Three-operation cycles also appear: +2, ×2, −1 gives 4, 6, 12, 11, 13, 26, 25. Divide the steps into groups of three.

When the constants change. Sometimes the cycle is fixed but the numbers inside it follow their own pattern: +17, ×7, +15, ×5, +13, ×3. The added numbers go 17, 15, 13 (down by 2) and the multipliers go 7, 5, 3 (down by 2).

Write the operation above each arrow. Compare every second step: odd steps share one operation, even steps share another.

What the exam asks. A jump too big for differences, with terms close to squares or cubes: 7, 26, 63, 124, ? or 0, 6, 24, 60, 120, ?. And, in the latest papers, a [New Pattern]: squares, cubes or products built on a Fibonacci-type list, such as 4, 9, 25, 64, 169, ? or 6, 15, 40, 104, 273, ?.

Why it works. The paper-setter starts from a simple base (1, 2, 3 … or 1, 2, 3, 5, 8 …), applies a power, and then nudges the result by a small amount. Strip off the nudge and the base shows itself. So look at the largest term first and ask: which square or cube is it near?

Position-based forms. The base is the position number n (1, 2, 3 …).

• n² − 1: 0, 3, 8, 15, 24, 35, 48
• n² + 1: 2, 5, 10, 17, 26, 37
• n² − n or n² + n: 0, 2, 6, 12, 20 or 2, 6, 12, 20, 30
• n³ − 1, n³ + 1: 0, 7, 26, 63 or 2, 9, 28, 65, 126
• n³ − n: 0, 6, 24, 60, 120, 210. Notice n³ − n = (n − 1) × n × (n + 1), the product of three consecutive numbers. So 210 = 5 × 6 × 7.

[[fig:0]]

To tell the forms apart, write the first three terms of each form. n² − 1 starts 0, 3, 8; n² + 1 starts 2, 5, 10; n² − n starts 0, 2, 6.

Squares of special bases. The bases need not be 1, 2, 3. They may be primes in order (the series 361, 289, 169, 121, 49 is 19², 17², 13², 11², 7², and the next base is the next prime, 5), or step by 2 (14², 16², 18², 20²).

Base is the previous term. In 1, 2, 5, 26, 677 each term is the previous term squared plus 1: 1² + 1 = 2, 2² + 1 = 5, 5² + 1 = 26, 26² + 1 = 677.

[New Pattern] Fibonacci-type roots. A Fibonacci-type list is one in which each number is the sum of the two before it: 1, 1, 2, 3, 5, 8, 13, 21, 34. Recent SSC papers build series on it in four ways.

• Cubes. 1, 8, 27, 125, 512 look unrelated, but the cube roots are 1, 2, 3, 5, 8. The next root is 5 + 8 = 13, so the next term is 13³ = 2197. The same idea with letters: A1, B1, C8, D27, E125, F512 carries the cubes of 1, 1, 2, 3, 5, 8 beside the letters, so the next term is G with 13³ = 2197.
• Squares. 4, 9, 25, 64, 169 are 2², 3², 5², 8², 13². The next root is 21, so the next term is 21² = 441.
• Products of neighbours. 6, 15, 40, 104, 273 are 2 × 3, 3 × 5, 5 × 8, 8 × 13, 13 × 21. The next term is 21 × 34 = 714.
• Roots that step by 2, or start from another pair. 1, 27, ?, 343 are 1³, 3³, ?, 7³, so the missing root is 5 and ? = 125. And 27, 64, 343, 1331 have the roots 3, 4, 7, 11, so the next root is 18 and the next term is 18³ = 5832.

[[fig:1]]

Steps. (1) Turn every term into its root, using the squares to 30 and the cubes to 15. If a term is not a power, try splitting it into two neighbouring numbers of the list. (2) Write the roots as a row and find the pattern: the sum of the two before, or a fixed step. (3) Extend the roots by one, then raise to the power or multiply. Add the roots first and cube afterwards; 1³ + 2³ = 9 is not 3³ = 27.

Last digit check: 26² ends in 6, so 26² + 1 ends in 7. Use the last digit to discard options at once.

What the exam asks. The terms rise fast but not by a fixed ratio, and the differences look like the series itself: 3, 4, 7, 11, 18, 29, ? or 1, 1, 2, 3, 5, 8, ?. The next term depends on more than the one term before it.

Why it works. In these series the pattern uses the last two terms (sometimes three) instead of only the last one. That is why single-step patterns such as + a or × b fail. It is also why the differences reproduce earlier terms: if each term is the sum of the two before it, then term − previous term equals the term before that.

In some series a term depends on more than the one term before it. Three such families come up again and again.

1. Fibonacci type: each term is the sum of the two before it. 2, 3, 5, 8, 13, 21. The series may start from any two numbers: 3, 4, 7, 11, 18, 29 is also Fibonacci type. The giveaway is that the differences reproduce the series itself: for 3, 4, 7, 11 the differences are 1, 3, 4, which are earlier terms. Whenever the differences look like the series shifted, think "sum of two previous terms".

[[fig:0]]

2. The 2n + 1 type. Each term is double the previous term plus 1: 1, 3, 7, 15, 31, 63. Starting from 1, the terms are one less than powers of 2. Similar forms are 2n − 1 or 2n + 2. These belong to the multiply-and-add family too, but they are worth knowing here because 2n + 1 = n + (n + 1): each term is the previous term added to itself, plus 1.

3. Sum of the squares of the digits. Each term is made from the digits of the previous term: square each digit and add. Start at 16: 1² + 6² = 37; 3² + 7² = 58; 5² + 8² = 89; 8² + 9² = 145; 1² + 4² + 5² = 42. Compute digit by digit, term by term.

If the differences repeat the earlier terms of the series, it is Fibonacci type. Check on three terms before you trust it.

A different starting pair does not change the idea. You still add the last two terms to get the next.

What the exam asks. Series that fit none of the earlier types. In recent papers they are mostly these: differences made of primes (10, 12, 15, 20, 27, 38, ? has differences 2, 3, 5, 7, 11; 89, 87, 84, 79, 72, ? falls by 2, 3, 5, 7); differences that are falling squares (99, 124, 140, 149, 153, ? adds 25, 16, 9, 4); series built from primes ± 1; decimals; and "which number is different" items on primes, such as 13, 17, 19, 23, 27.

Why it works. Primes and squares are famous lists, so they hide easily inside a difference row. The giveaway is a row that is irregular but not random: small numbers that never settle, yet each one is a prime or a square.

Some series belong to special families. When the usual tests fail, check these.

Primes. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The terms are the primes in order. A prime is a number with exactly two factors, so 9 (3 × 3), 15 (3 × 5), 21 (3 × 7) and 25 are not primes. Primes are not the same as odd numbers.

Prime ± 1 or prime × k. 4, 6, 8, 12, 14, 18 is each prime plus 1 (3, 5, 7, 11, 13, 17 plus 1). The differences (2, 2, 4, 2, 4) look irregular, so the difference method fails. This is the clue: a series with irregular differences that are always small is probably linked to primes.

[[fig:0]]

Decimals. To handle 0.2, 0.6, 1.2, 2.0, 3.0 multiply every term by 10: 2, 6, 12, 20, 30. The differences are 4, 6, 8, 10, so the next difference is 12 and the next term is 42, which is 4.2 after dividing by 10.

Digit-based items. Some series depend on digits: the sum of digits, the reversed number (13, 31, 17, 71), or the product of digits. Compute the quantity separately for each term and compare the results.

Repeated terms. A term may appear twice, as in 2, 2, 4, 4, 8, 8. Treat each pair as one step.

Do not hunt for a difference pattern when the differences are irregular. Check whether the terms themselves are primes or primes ± 1.

What the exam asks. Two forms. The classic one has no question mark: "find the wrong term" in a complete series such as 3, 7, 15, 31, 64, 127. The newer form asks which item does not follow the pattern or does not belong: 2, 4, 6, 8, 11 or ABC, DEF, GHI, JKL, JIK. In both, one item breaks a pattern that all the others obey.

Why it works. The setter builds the series correctly and then spoils one term. So the other terms agree with each other, and the spoiled one is the only one that disagrees. Find the agreement first, and the culprit is obvious.

In this type there is no question mark. The series is complete, but one term does not fit the pattern, and you must pick it out. The question reads "Find the wrong term" or "Select the incorrect number".

Method.
1. Find the pattern from the terms that agree with each other. The later, larger terms usually show the pattern best, so start from the end if the beginning looks odd.
2. Check each term against the pattern, from left to right.
3. The term that fails is the wrong one. The terms after it are built from the correct value.

Take 3, 7, 15, 31, 64, 127. The pattern 2n + 1 gives 3, 7, 15, 31, 63, 127. So 64 should have been 63.

[[fig:0]]

A check that never fails. When you suspect a term, test the next term against the "correct" value. Here 127 = 2 × 63 + 1, so the series was built from 63, and 64 is certainly the error. If the next term does not fit either, re-examine your pattern.

Where can the wrong term hide? At the very start, in the middle or at the very end. Do not check only the first few terms.

Pattern types. Any of the earlier types can be hidden here: differences, multiply-and-add, squares, interleaved. If the series has a clear alternating pattern, split it into odd and even places first.

The "does not belong" form. Recent papers also ask which number or group does not follow the pattern: 2, 4, 6, 8, 11 (all even except 11); ABC, DEF, GHI, JKL, JIK (each group is three consecutive letters in order, but JIK is not); AB12, CD34, 56EF, GH78 (the format "two letters, then two digits" breaks in 56EF). There is no calculation here. List what the majority share (parity, primes, an order, a format), and the item that breaks it is the answer.

First find the pattern from the clear terms. Then the wrong term is the one that breaks it, and the term after it confirms.

If you cannot find any pattern from the unbroken terms, write difference rows again: the error may have disturbed the first row.

What the exam asks. A row of letters with a question mark: A, E, I, M, ? or M, O, R, V, ? or B, D, G, K, ?. The options differ in the last letter, so only one letter matters. Recent papers also hide the same idea in a code table (A = Z, B = X, C = U …).

Why it works. Letters are positions 1 to 26 in disguise, arranged in a ring. Convert them to numbers and a letter series becomes a number series on the ring: a constant gap, a gap that grows, or a gap that cycles.

In the simplest letter series every letter is the same number of steps from the one before. In E, I, M, Q, each letter moves 4 steps ahead.

Method.
1. Convert every letter into its position (A = 1 … Z = 26).
2. Write the gap between neighbours.
3. If the gaps are equal, add the gap to the last position.
4. Convert back to a letter. If the answer goes above 26, subtract 26.

[[fig:0]]

Backward series (descending letters) have a negative gap. W, S, O, K are 23, 19, 15, 11, a gap of −4, so the next is 7 = G.

Wrap-around. Letters form a ring, so the gap can carry you across Z. In T, B, J, R the positions are 20, 2, 10, 18. The jump from 20 to 2 looks like −18, but it is really +8: 20 + 8 = 28, and 28 − 26 = 2. The next is 18 + 8 = 26 = Z.

Options first. Examiners usually change the last letter of the options. If the options end in different letters, you do not need the whole series: find the gap and test only the last letter.

Find the gap on two pairs at least. Convert to numbers, add, and bring the answer back into 1 to 26.

When the gap itself changes. The next step is the same method, one level deeper.

Here the gap between neighbouring letters is not the same each time. The gap itself follows a pattern: it may grow (+3, +4, +5, …) or repeat in a cycle (−3, −2, −1, −3, −2, …).

Method.
1. Convert to positions.
2. Write every gap, with its sign.
3. Study the row of gaps exactly as you would study a number series: is it constant, growing, alternating, a repeating cycle?
4. Extend the gap row by one, and apply it to the last position.

[[fig:1]]

Take C, F, J, O. Positions 3, 6, 10, 15, gaps 3, 4, 5. The next gap is 6, so the next position is 21 = U.

A repeating cycle of gaps. Z, W, U, T, Q has positions 26, 23, 21, 20, 17 and gaps −3, −2, −1, −3. The cycle −3, −2, −1 has begun again, so the next gap is −2 and the next letter is 15 = O.

Across Z. The gaps keep counting even when the series wraps. R, T, W, A, F: positions 18, 20, 23, 1, 6. The jump from 23 to 1 is not −22: it is 23 + 4 = 27, and 27 − 26 = 1, a gap of +4. So the gaps are 2, 3, 4, 5, and the next gap is 6, giving 12 = L.

Write every gap. If the gaps grow, extend them; if they repeat, extend the cycle.

Some gap rows can be read in two ways (for example 2, 3, 5, then 7 for primes, or 8 for a Fibonacci-style pattern). If two options both fit, the examiner usually intended the simpler one.

[New Pattern] Letter mappings. Some items give a code table instead of a row: A = Z, B = X, C = U, D = Q, find the code for E. Turn the codes into positions: 26, 24, 21, 17. The gaps are −2, −3, −4, so the next gap is −5 and the position is 12, which is L.

Shifting every letter of a word. Sometimes a whole word moves by the same gap. Change each letter of TOK, THE, HAS, ALL, COW to the next letter and you get UPL, UIF, IBT, BMM, DPX. Mark the vowels A, E, I, O, U in the new words: only BMM and DPX have none, so the answer is two. Note that T becomes U and H becomes I, which are vowels.

What the exam asks. Each term is a cluster of two or three letters, and you must give the next cluster or fill a blank in the middle: BD, FH, JL, NP, ? or PMT, OOS, NQR, MSQ, ? or BKS, DMU, ?, HQY, JSA. This is the most frequent letter-series type in recent papers.

Why it works. The setter runs one simple letter series in every column of the cluster, and the columns are independent: the first letters move on their own, the second letters on their own. So a cluster series is just two or three small letter series standing side by side.

Here each term is a cluster of two or three letters, like CX or ACF. Every position inside the cluster has its own pattern, and the positions move independently.

Method.
1. Take the first letters of all the clusters as one series; take the second letters as another series; and the third letters, if any, as a third.
2. Convert each series to positions and find its gap, with sign.
3. Apply each gap to the last cluster and build the next cluster.
4. Check the options by the position where they differ.

[[fig:0]]

CX, EV, GT, IR: first letters C, E, G, I are 3, 5, 7, 9 (+2). Second letters X, V, T, R are 24, 22, 20, 18 (−2). So the next cluster is K (11) and P (16): KP.

Same shift in all positions. BD, EG, HJ, KM: each letter moves +3, so the next is NP. ACF, DFI, GIL, JLO is the same idea with three letters.

Different shifts and signs. BYM, EVO, HSQ, KPS: the first letters go +3 (2, 5, 8, 11), the second go −3 (25, 22, 19, 16) and the third go +2 (13, 15, 17, 19). The next cluster is N (14), M (13), U (21).

Another kind. Sometimes each cluster is built from the previous cluster by a fixed transformation: the letters shift, or swap places, or reverse. Find the transformation from the first two clusters and test it on the third.

Treat every position as its own series, with its own gap and its own sign.

Look at the options before you start. Often only one position needs checking.

Blank in the middle. When the blank is inside the row (BKS, DMU, ?, HQY, JSA), work column by column and use both neighbours. The first letters B, D, ?, H, J step by 2, so the blank is F. The second letters K, M, ?, Q, S give O, and the third letters S, U, ?, Y, A give W. The cluster is FOW.

Opposite letters and vowels in a series.

Opposite letters. The alphabet read backwards is the mirror of the alphabet read forwards: A with Z, B with Y, C with X, and so on. The positions of opposite letters always add up to 27. So the opposite of a letter at position p is at position 27 − p (the opposite of G, position 7, is position 20, which is T).

A series may alternate between a letter and a position from the other end. A, Z, B, Y, C, X is really two series woven together: the odd places go forward (A, B, C, D) and the even places go backward (Z, Y, X, W).

[[fig:1]]

Method.
1. Split the terms into odd and even places.
2. Solve each as its own series.
3. Decide which line holds the missing term, and answer from that line.

Take A, Z, B, Y, C, X, ?. The seventh place is odd, so the answer is D. It is not W; W would be the eighth term.

Another form. D, W, G, T, J, Q has odd places D, G, J (+3 each: 4, 7, 10) and even places W, T, Q (23, 20, 17: −3 each). Each even letter is also the opposite of the letter before it (4 + 23 = 27, 7 + 20 = 27, 10 + 17 = 27). The next term is at an odd place: 10 + 3 = 13 = M.

Vowels. The five vowels are A, E, I, O, U. Vowel series run in this order and then start again: U, O, I, E, A is the order backwards. Be careful when only vowels are given in a series such as A, E, I, O: a series with gaps 4, 4, 6 also fits. Let the options decide.

Opposite letters add up to 27. Mark the odd and even places before choosing the track.

Vowel clusters. AEI, EIO, IOU, ? drops the first vowel and adds the next one in the cycle A, E, I, O, U, A …, so the next cluster is OUA. To count vowels from the right, go U (1st), O (2nd), I (3rd), E (4th), A (5th).

Opposite-letter pairs as clusters. In ZA, BY, XC, DW, ? every cluster is a pair of opposite letters (26 + 1, 2 + 25, 24 + 3, 4 + 23, all 27). The first letters Z, B, X, D split into two tracks (Z, X, V going down by 2 and B, D going up by 2), so the next cluster starts with V, and its partner is 27 − 22 = 5, which is E: VE.

What the exam asks. Each term is a letter group with a number: ADG 3, EHK 6, ILO 9, MPS 12, ? or PHX 80, TLB 73, XPF 66, BTJ 59, ?. Recent NTPC papers use three-letter groups with a number, and ask for the next group.

Why it works. The setter runs three small letter series (one per column) and one number series side by side. They are independent, or the number is tied to the letters. So the question is really four small series. The skill is to solve only as many of them as the options force you to.

Each term has a letter part and a number part, like B5 or 12C3. The letters form one series and the numbers form another, and often they move at different speeds, in different directions.

Method.
1. Separate the letter row from the number row.
2. Convert the letters to positions.
3. Solve each row as its own series (constant gap, variable gap, ×, squares …).
4. Join the two answers.

[[fig:0]]

Take B5, D9, F13, H17. Letters B, D, F, H are +2 each, so the next is J. Numbers 5, 9, 13, 17 are +4 each, so the next is 21. Answer J21.

Rows with opposite directions. 42E, 36H, 30K, 24N: the numbers fall by 6 while the letters rise by 3 (E, H, K, N). The next term is 18Q.

The number belongs to the letter. In C9, F18, I27, L36, the letters are C (3), F (6), I (9), L (12), and the number is 3 times the position: 3 × 3 = 9, 3 × 6 = 18, 3 × 9 = 27, 3 × 12 = 36. The next term is O (15) with 3 × 15 = 45, so O45. Always check this link before you treat the two rows separately.

Clusters with a number. A term like ECP 1 or LJL 62 combines a letter cluster with a number. Solve the cluster series by position and the numbers as a separate number series.

Let the options choose the part. In the PHX example the options carry 44, 49, 50 and 52. The number row 80, 73, 66, 59 falls by 7, so the next number is 52 and only one option has it. No letter work is needed. When every option carries the same number (ADG 3 … ? 15), the letters must decide. Solve the part in which the options differ.

The number has its own pattern. It may fall by a fixed step (80, 73, 66 …), rise by a fixed step (2, 5, 8, 11), or be the square of the step number (1, 4, 9, 16, 25).

The three letter columns often move alike. In PHX, TLB, XPF, BTJ every letter goes up by 4 (P, T, X, B, F and H, L, P, T, X and X, B, F, J, N). Check the wrap across Z with the 26 step.

Never apply one pattern to both parts. Separate the rows, then check whether the number is built from the letter.

What the exam asks. A string of letters with several underscores, and four options that list the missing letters in order: uw_wuv_vuwu_uvu_ or ab_cb_bn_b__n_b. These appear in Steno papers in particular.

Why it works. The string is made by repeating a short block again and again, and some places are hidden. Once you know the block, every blank can be read from the same place of another block.

In this type a block of letters repeats again and again, but some places are empty. You are asked to fill the blanks, and the options list the letters for the blanks, in order.

p q _ r p q _ r p q _ r is built from the block "p q s r" repeated three times. Once you know the block, every blank is filled from the same place in another block.

Method.
1. Count all the places, including the blanks.
2. Find a block length that divides the count (16 places can be 2 × 8, 4 × 4 or 8 × 2; try 3, 4, 5, 6 as block lengths).
3. Write the string in rows of that length, one block per row. The same column now holds the same letter.
4. Fill each blank from the other rows in its column.
5. Write the blanks in order and match with the options.

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If the rows do not line up (the same column shows two different letters), the block length is wrong. Try the next length.

When the block is not fixed. Some items grow: a, bb, aa, bbb, aaa, bbbb. There is no fixed block. Look at the sizes of the groups (1, 2, 2, 3, 3, 4) and the letters alternate. Fill the blanks using the group sizes.

Blocks of 5 or 8. The block need not be 3 or 4 long. In ab_cb_bn_b__n_b there are 15 places, and the block is 5: abncb, abncb, abncb. In uw_wuv_vuwu_uvu_ there are 16 places; blocks of 4 do not agree, but every second row does (uwuw, uvuv, uwuw, uvuv), so the block is 8 long.

Count the blanks as places. A block length that does not divide the total number of places is wrong.

Shortcut with the options. Fill the blanks you are sure of, then see which options still agree with them.