Number Series, Wrong-Term Series and Equation Comparison

This chapter answers the Number Series and Quadratic Equations items of the Quantitative Aptitude section of the Prelims of IBPS Clerk, SBI Clerk, IBPS PO, SBI PO and IBPS RRB, together with the 'solve I and II and compare x and y' and 'Quantity I and Quantity II' items that follow the same method. It teaches how to read a series by differences, ratios, multiply-and-add rules, squares, cubes and primes; how to find the wrong term; how to solve a quadratic by splitting the middle term and read the signs of its roots; and how to compare two sets of roots and choose among the five standard relations, including the final option, 'x = y or the relation cannot be established'. Every question is numerical, and a wrong answer costs one-fourth of a mark.

Reading a series: the standard patterns

A series is built from one rule applied again and again. Start by looking at the size of the jump: a slow rise points to differences, a rapid rise to ratios or multiply-and-add, and terms close to 16, 25, 36, 64, 125, 216 point to squares and cubes. Always compute the first differences first; if they are not constant, compute the differences of the differences.

The families of rules
PatternExampleHow to spot it
Constant difference7, 12, 17, 22Differences all 5
Differences in arithmetic progression5, 6, 9, 14, 21Differences 1, 3, 5, 7; second difference 2
Differences are squares3, 4, 8, 17, 33, 58Differences 1, 4, 9, 16, 25
Multiply and add: × k + c3, 7, 15, 31, 63Each term is twice the last plus 1
Growing multipliers2, 3, 8, 27, 112× 1 + 1, × 2 + 2, × 3 + 3, × 4 + 4
Squares ± k5, 10, 17, 26, 372² + 1, 3² + 1, 4² + 1, 5² + 1, 6² + 1
Cubes ± k9, 28, 65, 1262³ + 1, 3³ + 1, 4³ + 1, 5³ + 1
Primes or their squares4, 9, 25, 49, 1212², 3², 5², 7², 11²
🧠 Memory hook: size, then differences, then ratios
Ask three questions in this order, each taking under five seconds: how fast does it grow, are the differences simple, are the ratios near a whole number. If two interleaved patterns are present (terms at odd places behave unlike those at even places), split the series into two and treat each separately.

Missing-term series: worked examples

Example 1 (differences). 6, 7, 11, 20, 36, ? The differences are 1, 4, 9, 16, the squares of 1, 2, 3, 4, so the next difference is 5² = 25 and the missing term is 36 + 25 = 61.

Example 2 (multiply and add). 2, 3, 8, 27, 112, ? The rise is quick, so test × k + c. 2 × 1 + 1 = 3, 3 × 2 + 2 = 8, 8 × 3 + 3 = 27, 27 × 4 + 4 = 112. The multiplier and the added number both grow by 1, so the next term is 112 × 5 + 5 = 565.

Example 3 (two interleaved series). 4, 21, 6, 18, 8, 15, ? Odd places give 4, 6, 8 (adding 2) and even places give 21, 18, 15 (subtracting 3). The seventh term continues the odd-place series, so it is 10.

Example 4 (cubes with a changing add-on). 2, 12, 36, 80, 150, ? Each term is n³ + n²: 1 + 1, 8 + 4, 27 + 9, 64 + 16, 125 + 25. The next is 216 + 36 = 252. Note that this equals n²(n + 1): when terms look like a product of a square and a neighbour, factorise them.

⚠️ Two rules can fit the first terms
3, 6, 12 fits both × 2 and the differences 3, 6. Never fix the rule from three terms. Test it on every given term, and prefer the rule that explains all of them with the simplest numbers. A rule that needs a different exception at one place is almost always the wrong rule.

Wrong-term series

In a wrong-term question one term does not obey the rule that all the others obey, and you must name it or give its correct value. The setter builds a perfect series and then replaces one term. So work from the terms you trust: the first few and the last two or three.

  • Step 1: find the rule from the two ends of the series, where an altered term is less likely to mislead you.
  • Step 2: run the rule term by term. The wrong term is the one that fails to give its successor and is not given by its predecessor.
  • Step 3: put back the correct value and confirm every term now fits.

Example A. 5, 11, 23, 48, 95, 191. Each good term is the last × 2 + 1 (5 → 11 → 23 → 47 → 95 → 191). From 23 the rule gives 47, not 48, and 95 is exactly 47 × 2 + 1, so 48 is wrong and 47 is correct.

Example B. 4, 5, 9, 19, 34, 59. The differences should be the squares 1, 4, 9, 16, 25: 4 → 5 → 9 → 18 → 34 → 59. The term 19 gives differences 10 and 15, so 19 is wrong and should be 18.

🧠 Parity, divisibility and size are quick detectors
In a series built by × 2 + 1 from an odd start, every term is odd, so an even term is wrong at once. In a series of multiples of 3, a term that is not divisible by 3 is the culprit. In a series that rises steadily, a term that jumps or falls out of line stands out. Use these before any arithmetic.
⚠️ In a recurrence the next term follows the correct one
When each term is built from the previous one, the term after the wrong one still follows the original, correct value, not the printed one. So if the successor fits your guess for the corrected term, that confirms the guess. Do not blame the successor merely because it fails the rule from the printed wrong term.

Quadratics: factorise, read the signs, list the roots

To solve ax² + bx + c = 0 by splitting the middle term, find two numbers whose product is a × c and whose sum is b. For x² − 7x + 12 = 0, the numbers −3 and −4 have product 12 and sum −7, so (x − 3)(x − 4) = 0 and x = 3 or 4. For 2x² − 11x + 12 = 0, the product is 24 and the sum −11, so −3 and −8: 2x² − 8x − 3x + 12 = 2x(x − 4) − 3(x − 4) = (2x − 3)(x − 4), giving x = 1.5 or 4.

Signs of the roots from the signs of b and c (a > 0)
CaseWhat it meansExample and roots
c > 0, b > 0Both roots negativex² + 9x + 20 = 0: −4, −5
c > 0, b < 0Both roots positivex² − 9x + 20 = 0: 4, 5
c < 0Roots of opposite signs; the larger in size has the sign opposite to that of bx² − 5x − 14 = 0: 7, −2
🎯 Why the signs follow
If the roots are p and q, the equation is x² − (p + q)x + pq = 0. So c = pq decides whether the roots share a sign, and −b = p + q gives their common sign when they do. This lets you reject a wrong factor pair at once.

Comparing x and y and the five relations

In 'I. … II. …' items, solve I for x and II for y, write each as a set of values on a number line, and test every pair (x, y). A relation can be chosen only if it holds for all pairs. The five options are x > y, x ≥ y, x < y, x ≤ y and 'x = y or the relation cannot be established'.

  • x > y holds when the smallest x exceeds the largest y; x ≥ y when the smallest x is at least the largest y.
  • x < y holds when the largest x is below the smallest y; x ≤ y when the largest x is at most the smallest y.
  • If none of the four holds for all pairs, the answer is the fifth option. It is also the answer when x and y are single, equal values.

Example 1. I. x² − 7x + 12 = 0, II. y² − 9y + 20 = 0. Here x ∈ {3, 4} and y ∈ {4, 5}. The pairs give 3 < 4, 3 < 5, 4 = 4 and 4 < 5, so x is never greater and sometimes equal: x ≤ y. Example 2. I. x² − 5x − 14 = 0 gives x ∈ {7, −2}; II. y² + 3y − 10 = 0 gives y ∈ {2, −5}. Then 7 > 2 but −2 < 2, so no relation holds for all pairs: cannot be established.

⚠️ Touching sets: ≥ versus >
When the smallest x equals the largest y, the correct relation is the weak one (≥ or ≤), not the strict one. If x ∈ {1.5, 4} and y ∈ {1, 1.5}, the pairs give 1.5 > 1, 1.5 = 1.5, 4 > 1, 4 > 1.5, so the answer is x ≥ y. Marking x > y overlooks the single equal pair, and marking 'x = y or cannot be established' overlooks that every pair obeys x ≥ y.

Linear pairs, powers and Quantity I versus II

Not every item is a quadratic. A pair of linear equations in x and y gives one value of each, so the answer is one of >, < or (rarely) =. For I. 3x + 2y = 22 and II. 2x + 5y = 33, multiply I by 5 and II by 2: 15x + 10y = 110 and 4x + 10y = 66; subtracting gives 11x = 44, so x = 4 and y = 5, hence x < y. Equations of the form x² = k give two roots, ±√k, and y³ = k gives one real value, so x² = 225 gives x ∈ {15, −15} while y³ = 3375 gives y = 15.

In Quantity I and Quantity II items, each quantity is a number found from a short calculation (a root, a sum of roots, a percentage), and the same five-way logic applies to the two numbers. Example. Quantity I: the sum of the roots of 2x² − 11x + 12 = 0. Quantity II: the larger root of y² − 7y + 10 = 0. The roots of the first are 1.5 and 4, whose sum is 5.5; the roots of the second are 2 and 5, so the larger is 5. Since 5.5 > 5, Quantity I > Quantity II. A shortcut for the sum is −b/a = 11/2 = 5.5.

⚠️ Read what the question asks of each root
The sum of the roots (5.5), the larger root (4) and the smaller root (1.5) of the same equation are three different quantities. Also, in a cube-root item the real value is single, but in a square item both signs count. In the exam the quickest route is to list the roots of each equation first and read the question afterwards.

Key takeaways

  • Read a series in order: size of growth, first differences, second differences, ratios, then × k + c, squares, cubes and primes; split interleaved series into two.
  • Never fix the rule from three terms; test it on every term and choose the simplest rule that explains them all.
  • For a wrong term, find the rule from the two ends, run it term by term, and use parity, divisibility and size to spot the culprit quickly.
  • Solve a quadratic by splitting the middle term: two numbers with product a × c and sum b; the signs of c and b give the signs of the roots.
  • A relation between x and y is valid only if it holds for every pair of roots; if none of the four does, the answer is 'x = y or the relation cannot be established'.
  • When the nearest values touch, choose the weak relation (≥ or ≤); remember x² = k has two roots, y³ = k only one, and the sum of the roots is −b/a.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Find the missing term: 6, 13, 27, 55, 111, ?

    1. 221
    2. 223
    3. 222
    4. 233
    Show answer

    Answer: B — 223

    Each term is the previous term × 2 + 1: 6 → 13 → 27 → 55 → 111 → 223. A reply of 222 treats the rule as plain doubling and forgets the + 1.

  2. Find the missing term: 12, 13, 17, 26, 42, ?

    1. 63
    2. 65
    3. 71
    4. 67
    Show answer

    Answer: D — 67

    The differences are 1, 4, 9, 16, the squares of 1 to 4, so the next difference is 25 and the term is 42 + 25 = 67. The value 65 comes from adding 23, which breaks the square pattern.

  3. Find the missing term: 2, 12, 36, 80, 150, ?

    1. 252
    2. 240
    3. 216
    4. 262
    Show answer

    Answer: A — 252

    The terms are n³ + n² for n = 1 to 5 (1 + 1, 8 + 4, 27 + 9, 64 + 16, 125 + 25), so the next is 6³ + 6² = 216 + 36 = 252. The value 216 is only the cube and leaves out the square.

  4. In the series 3, 20, 5, 17, 7, 14, 9, ? two patterns alternate. The eighth term is

    1. 12
    2. 10
    3. 11
    4. 13
    Show answer

    Answer: C — 11

    Odd places 3, 5, 7, 9 rise by 2; even places 20, 17, 14 fall by 3, so the eighth term is 14 − 3 = 11. The value 13 continues the rise of 2 in the wrong series.

  5. Find the missing term: 4, 9, 25, 49, 121, ?

    1. 144
    2. 169
    3. 196
    4. 225
    Show answer

    Answer: B — 169

    These are the squares of consecutive primes 2, 3, 5, 7, 11, so the next is 13² = 169. The value 144 is 12², but 12 is not prime, so the pattern breaks.

  6. Exactly one term is wrong in the series 7, 15, 31, 64, 127, 255. Which term is wrong?

    1. 64
    2. 31
    3. 127
    4. 15
    Show answer

    Answer: A — 64

    Each good term is the last × 2 + 1, so the terms should be 7, 15, 31, 63, 127, 255. Thus 64 is wrong, and it is even although every term of this rule is odd. The term 127 is correct because 63 × 2 + 1 = 127.

  7. One term of 4, 6, 12, 30, 92, 315 is wrong. What should replace it?

    1. 80
    2. 100
    3. 96
    4. 90
    Show answer

    Answer: D — 90

    The multipliers are 1.5, 2, 2.5, 3, 3.5: 4 → 6 → 12 → 30 → 90 → 315. The term 92 is not 30 × 3 and is not 315 ÷ 3.5, so it should be 90. The value 96 is 30 × 3.2, which is not in the pattern.

  8. Exactly one term is wrong in the series 11, 13, 17, 23, 31, 40, 53. Which is it?

    1. 17
    2. 40
    3. 31
    4. 23
    Show answer

    Answer: B — 40

    The differences should rise by 2 each time: 2, 4, 6, 8, 10, 12, giving 11, 13, 17, 23, 31, 41, 53. So 40 should be 41. The value 31 is consistent with both its neighbours' differences 8 and 10.

  9. Statements on finding a wrong term. P: In a series built by × 2 + 1 from an odd first term, an even term must be wrong. Q: The wrong term is always one of the middle terms. R: A term that neither follows from its predecessor nor leads to its successor under the rule is the suspect. S: After correcting the suspect, every term must fit one rule. Which statement is NOT correct?

    1. P
    2. R
    3. Q
    4. S
    Show answer

    Answer: C — Q

    Q is false: the altered term can be the first, the last or any in between, so the ends must be checked as well. P, R and S are all sound tests. P is the tempting doubt because of its absolute wording, but it is correct for this rule since odd × 2 + 1 is odd.

  10. I. x² − 7x + 12 = 0.
    II. y² − 9y + 20 = 0. Solve both equations and compare x and y.

    1. x > y
    2. x < y
    3. x = y or the relation cannot be established
    4. x ≤ y
    Show answer

    Answer: D — x ≤ y

    I gives (x − 3)(x − 4) = 0, so x = 3 or 4; II gives (y − 4)(y − 5) = 0, so y = 4 or 5. Every pair has x ≤ y, with equality at 4 and 4, so x < y fails and x ≤ y holds.

  11. I. x² − 5x − 14 = 0.
    II. y² + 3y − 10 = 0. Solve both equations and compare x and y.

    1. x = y or the relation cannot be established
    2. x > y
    3. x < y
    4. x ≥ y
    Show answer

    Answer: A — x = y or the relation cannot be established

    I gives (x − 7)(x + 2) = 0, so x = 7 or −2; II gives (y + 5)(y − 2) = 0, so y = −5 or 2. Then 7 > 2 but −2 < 2, so no relation holds for all pairs. The tempting x > y reflects only the larger roots.

  12. I. 2x² − 11x + 12 = 0.
    II. 2y² − 5y + 3 = 0. Solve both equations and compare x and y.

    1. x > y
    2. x ≥ y
    3. x < y
    4. x ≤ y
    Show answer

    Answer: B — x ≥ y

    I gives (2x − 3)(x − 4) = 0, so x = 1.5 or 4; II gives (2y − 3)(y − 1) = 0, so y = 1.5 or 1. The pairs give 1.5 = 1.5 and every other pair has x > y, so x ≥ y holds. The strict x > y fails at the equal pair.

  13. I. x² + 9x + 20 = 0.
    II. y² + 13y + 42 = 0. Solve both equations and compare x and y.

    1. x < y
    2. x ≤ y
    3. x > y
    4. x = y or the relation cannot be established
    Show answer

    Answer: C — x > y

    I gives (x + 4)(x + 5) = 0, so x = −4 or −5; II gives (y + 6)(y + 7) = 0, so y = −6 or −7. Both are negative, and the smallest x, −5, exceeds the largest y, −6, so x > y. The tempting x < y comes from comparing sizes instead of values.

  14. I. 3x + 2y = 22.
    II. 2x + 5y = 33. Solve the pair and compare x and y.

    1. x > y
    2. x < y
    3. x ≥ y
    4. x = y or the relation cannot be established
    Show answer

    Answer: B — x < y

    Multiplying I by 5 and II by 2 gives 15x + 10y = 110 and 4x + 10y = 66, so 11x = 44 and x = 4; then y = 5. Hence x < y. The relation x > y comes from swapping the two values.

  15. I. x² − 225 = 0.
    II. y³ = 3375. Solve both equations and compare x and y.

    1. x ≤ y
    2. x > y
    3. x < y
    4. x ≥ y
    Show answer

    Answer: A — x ≤ y

    I gives x = 15 or −15; II gives the single real value y = 15 since 15³ = 3375. The pairs give 15 = 15 and −15 < 15, so x ≤ y holds, while x < y fails at the equal pair and x ≥ y fails at −15.

  16. Quantity I: the sum of the roots of 2x² − 11x + 12 = 0. Quantity II: the larger root of y² − 7y + 10 = 0. Compare the two quantities.

    1. Quantity I < Quantity II
    2. Quantity I ≤ Quantity II
    3. Quantity I = Quantity II or the relation cannot be established
    4. Quantity I > Quantity II
    Show answer

    Answer: D — Quantity I > Quantity II

    The roots of the first equation are 1.5 and 4, so the sum is 5.5 (also −b/a = 11/2). The second equation gives 2 and 5, so the larger root is 5. Since 5.5 > 5, Quantity I is greater. Confusing the sum with the larger root 4 would reverse the result.