◆ QA · Algebra

Progressions & Series , formulas + CAT PYQs

Focused Algebra kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Progressions & Series is here.

49CAT PYQs
Algebrachapter
Formulas PYQs (49)↩ Full Algebra chapter

Algebra, formula sheet

Show the full Algebra formula sheet (explanations + basic examples)
1Polynomials & zeroes
  • Plain English: a polynomial is just a sum of x-powers; its "zeroes" are the x-values that make it equal 0.
  • A polynomial of degree n: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
  • k is a zero of p(x) if p(k) = 0. Zeroes are the x-coordinates where y = p(x) cuts the x-axis.
  • Max zeroes = degree: linear → 1, quadratic → 2, cubic → 3, degree n → n.
  • e.g. p(x) = x² − 9 has degree 2, so at most 2 zeroes: x = 3 and x = −3.
2Sum & product of roots
  • Plain English: you can read the sum and product of the roots straight off the coefficients, no need to solve.
  • Quadratic ax²+bx+c: α + β = −b/a, αβ = c/a
  • Cubic ax³+bx²+cx+d: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
  • Build the equation: x² − (sum)x + (product) = 0.
  • e.g. x² − 5x + 6 = 0: sum = 5, product = 6 ⇒ roots 2 and 3 (2+3=5, 2×3=6).
3Discriminant & nature of roots
  • Plain English: the discriminant D is a single number that tells you how many real roots a quadratic has, before solving.
  • For ax²+bx+c (a≠0): D = b² − 4ac
  • D > 0 → two distinct real roots; D = 0 → equal real roots; D < 0 → no real roots (complex).
  • D a perfect square (a,b,c rational) → roots are rational.
  • Roots: x = (−b ± √D)/2a
  • e.g. x² + x + 1: D = 1 − 4 = −3 < 0 ⇒ no real roots.
4Sum of squares of roots (trick)
  • Plain English: you can get α²+β² from the sum and product alone, never solve for the roots first.
  • α² + β² = (α+β)² − 2αβ
  • To minimise a sum-of-squares-of-roots expression in a parameter, complete the square, minimum is at the vertex.
  • e.g. if α+β = 3 and αβ = 2, then α²+β² = 9 − 4 = 5.
5Algebraic identities
  • Plain English: memorised expand/factor templates that turn ugly expressions into products (or vice-versa) instantly.
  • (a±b)² = a² ± 2ab + b²
  • a² − b² = (a+b)(a−b)
  • (a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
  • a³ ± b³ = (a±b)(a² ∓ ab + b²)
  • a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca)
  • e.g. 97×103 = (100−3)(100+3) = 100² − 3² = 9991.
6Linear equations in two variables
  • Plain English: comparing the coefficient ratios tells you whether two lines cross once, never, or lie on top of each other.
  • a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
  • Unique solution (intersecting): a₁/a₂ ≠ b₁/b₂
  • No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  • Infinite solutions (coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
  • e.g. x+y=2 and 2x+2y=5: ratios 1/2 = 1/2 ≠ 2/5 ⇒ parallel, no solution.
7Inequalities, basic rules
  • Plain English: inequalities behave like equations, except multiplying or dividing by a negative reverses the arrow.
  • Adding/subtracting keeps direction; multiplying by a negative flips the sign.
  • If X > Y > 0 then 1/X < 1/Y.
  • For x > 0: x + 1/x ≥ 2 (equality at x = 1).
  • e.g. −2x > 6 ⇒ divide by −2 and flip ⇒ x < −3.
8Quadratic inequalities
  • Plain English: factor it, then "< 0" means between the roots and "> 0" means outside the roots.
  • (x−m)(x−n) < 0, n > m ⇒ m < x < n (between the roots).
  • (x−m)(x−n) > 0 ⇒ x < m or x > n (outside the roots).
  • Sign-of-product / wavy-curve method handles higher degree.
  • e.g. x² − 5x + 6 < 0 ⇒ (x−2)(x−3) < 0 ⇒ 2 < x < 3.
9Modulus (absolute value)
  • Plain English: |x| is the distance of x from 0, so it strips the sign and is never negative.
  • |x| = max(x, −x); −|x| ≤ x ≤ |x|.
  • |a+b| ≤ |a|+|b| and |a|−|b| ≤ |a−b|; |ab| = |a||b|.
  • |x| ≤ k ⇒ −k ≤ x ≤ k. |x| ≥ k ⇒ x ≥ k or x ≤ −k.
  • |f| + |g| = |f+g| only when f, g have the same sign.
  • e.g. |x| ≤ 3 ⇒ −3 ≤ x ≤ 3; |x − 4| = 2 ⇒ x = 6 or x = 2.
10AM-GM-HM inequality
  • Plain English: for positive numbers the plain average is always ≥ the geometric average, the go-to tool for "find the minimum".
  • For positive reals: AM ≥ GM ≥ HM, equality when all equal.
  • Two numbers: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).
  • AM × HM = GM²
  • If a₁a₂…aₙ = 1 then a₁+a₂+…+aₙ ≥ n.
  • e.g. for a = 2, b = 8: AM = 5 ≥ GM = √16 = 4. ✓
11Maxima & minima of a quadratic
  • Plain English: a parabola's turning point is at x = −b/2a; that's where the min (opens up) or max (opens down) lives.
  • ax²+bx+c: vertex at x = −b/2a; extreme value = (4ac − b²)/4a = −D/4a.
  • a > 0 → opens up → minimum; a < 0 → opens down → maximum.
  • min/max of max-of-two / min-of-two lines occurs where the two graphs intersect.
  • e.g. x² − 6x + 5: vertex at x = 3, minimum value = 9 − 18 + 5 = −4.
12Functions, domain, range, even/odd
  • Plain English: domain is what you may feed in, range is what comes out; even/odd describe the graph's symmetry.
  • Domain = allowed inputs; range = resulting outputs.
  • Even: f(−x) = f(x) (graph symmetric about y-axis), e.g. x², |x|.
  • Odd: f(−x) = −f(x) (symmetric about origin), e.g. x³, 1/x.
  • Inverse exists only if f is one-to-one.
  • e.g. f(x) = x³ is odd: f(−2) = −8 = −f(2). ✓
13Functional equations
  • Plain English: the form of a functional rule reveals the function, "turns + into ×" means exponential, etc.
  • f(x+y) = f(x)·f(y) ⇒ exponential type, f(x) = aˣ.
  • f(xy) = f(x)·f(y) ⇒ power/multiplicative; f(1) = 1.
  • If f(a+x) = f(a−x), the graph is symmetric about x = a; roots pair around a (sum of 4 roots = 4a).
  • e.g. f(x+y) = f(x)f(y) with f(1) = 3 ⇒ f(2) = f(1)² = 9.
14Graph shifting
  • Plain English: changes outside f() move the graph vertically; changes inside f() move it horizontally (and oppositely).
  • f(x)+c → shift up c; f(x)−c → shift down c.
  • f(x+c) → shift left c; f(x−c) → shift right c.
  • −f(x) → reflect in x-axis; f(−x) → reflect in y-axis.
  • e.g. y = (x−2)² is y = x² shifted 2 units right.
15Logarithm, definition
  • Plain English: log_b x just asks "what power of b gives x?", it's the inverse of raising to a power.
  • y = log_b x ⇔ x = bʸ (b > 0, b ≠ 1, x > 0).
  • log_a a = 1; log_a 1 = 0; a^(log_a m) = m.
  • e.g. log₂8 = 3 because 2³ = 8.
16Logarithm laws
  • Plain English: logs turn multiplication into addition, division into subtraction, and powers into multipliers.
  • log_a(xy) = log_a x + log_a y
  • log_a(x/y) = log_a x − log_a y
  • log_a(xᵐ) = m·log_a x
  • log_(aⁿ)(xᵐ) = (m/n)·log_a x
  • Change of base: log_a x = (log x)/(log a); log_a x = 1/log_x a
  • e.g. log₂40 = log₂(8×5) = log₂8 + log₂5 = 3 + log₂5.
17Indices (laws of exponents)
  • Plain English: same base, add exponents when multiplying, subtract when dividing, multiply when raising a power to a power.
  • pᵐ·pⁿ = pᵐ⁺ⁿ; pᵐ/pⁿ = pᵐ⁻ⁿ; (pᵐ)ⁿ = pᵐⁿ
  • pⁿ·qⁿ = (pq)ⁿ; (p/q)ⁿ = pⁿ/qⁿ
  • p⁻ⁿ = 1/pⁿ; p⁰ = 1; p^(1/n) = ⁿ√p
  • e.g. 2³·2⁴ = 2⁷ = 128; 8^(2/3) = (∛8)² = 2² = 4.
18Surds & rationalisation
  • Plain English: a surd is an unresolved root like √2; "rationalising" clears it from a denominator using the conjugate.
  • √(ab) = √a·√b; √(a/b) = √a/√b.
  • Rationalise a/(b+√c) by multiplying top & bottom by the conjugate (b−√c).
  • If a+√b is a root of a rational quadratic, so is its conjugate a−√b.
  • e.g. 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/2.
19Arithmetic Progression (AP)
  • Plain English: an AP adds the same step d each time; its sum is just "how many terms × the average of first and last".
  • Constant difference d. nth term: Tₙ = a + (n−1)d
  • Sum: Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last)
  • Arithmetic mean of a, b: A = (a+b)/2. Middle term = average of an odd count of AP terms.
  • e.g. 2, 5, 8, …: T₄ = 2 + 3×3 = 11; sum of first 4 = 4/2·(2+11) = 26.
20Geometric Progression (GP)
  • Plain English: a GP multiplies by the same ratio r each time; if |r| < 1 the infinite sum settles to a finite value.
  • Constant ratio r. nth term: Tₙ = a·rⁿ⁻¹
  • Sum: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
  • Infinite sum (|r| < 1): S∞ = a/(1 − r)
  • Geometric mean: G = √(ab).
  • e.g. 1 + ½ + ¼ + … = 1/(1 − ½) = 2.
21Harmonic Progression (HP)
  • Plain English: an HP is just an AP flipped, take reciprocals and you're back to a normal AP.
  • a, b, c… in HP ⇔ 1/a, 1/b, 1/c… in AP.
  • Harmonic mean of a, b: H = 2ab/(a+b)
  • nth term of HP = 1/(nth term of the corresponding AP).
  • e.g. 1, ½, ⅓, ¼ is an HP (reciprocals 1, 2, 3, 4 form an AP).
22Standard summation formulas
  • Plain English: ready-made closed forms for adding up the first n numbers, their squares, and their cubes.
  • Σn = n(n+1)/2
  • Σn² = n(n+1)(2n+1)/6
  • Σn³ = [n(n+1)/2]²
  • Telescoping: 1/(k·(k+1)) = 1/k − 1/(k+1).
  • e.g. 1 + 2 + … + 10 = 10×11/2 = 55.
23Common terms of two APs
  • Plain English: numbers shared by two APs themselves form an AP whose step is the LCM of the two steps.
  • Common terms of two APs form a new AP with common difference = LCM of the two differences.
  • Find the first common term, then count multiples of the LCM up to the smaller upper limit.
  • e.g. 2,5,8,… and 3,7,11,…: first common term 11, new step = LCM(3,4) = 12 ⇒ 11, 23, 35, …
24Recurrence & tₙ from Sₙ
  • Plain English: if you know the running total Sₙ, each term is just this total minus the previous total.
  • If Sₙ given: aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).
  • Alternating-sum sequences: subtract consecutive defining equations to isolate a term.
  • e.g. Sₙ = n² ⇒ a₅ = S₅ − S₄ = 25 − 16 = 9.
25Integer / Diophantine solutions
  • Plain English: once you spot one whole-number solution, all the rest come by stepping x and y in fixed jumps.
  • ax + by = c with one integer solution (x₀, y₀): all others are x₀ + (b/g)t, y₀ − (a/g)t, where g = gcd(a,b).
  • Bound the count using the given ranges on x and y.
  • e.g. 2x + 3y = 12: (x,y) = (3,2) works; next is (0,4), then (6,0), x jumps by 3, y by 2.
26When does Aᴮ = 1?
  • Plain English: a power equals 1 in exactly three situations, check all three or you'll miss cases.
  • Base = 1 (any exponent), or
  • Exponent = 0 (base ≠ 0), or
  • Base = −1 with an even exponent.
  • e.g. (−1)⁴ = 1 (base −1, even power); 7⁰ = 1 (zero power); 1⁹⁹ = 1 (base 1).
27Three terms in AP / GP
  • Plain English: centering three terms on a middle value makes their sum (AP) or product (GP) collapse to one symbol.
  • Three in AP: take a−d, a, a+d (their sum = 3a).
  • Three in GP: take a/r, a, ar (product = a³).
  • Three consecutive integers as roots: n−1, n, n+1.
  • e.g. three numbers in AP summing to 18 ⇒ middle = 6, so 6−d, 6, 6+d.
28|x − a| as distance (modulus sums)
  • Plain English: read |x−a| as "distance from a", and sums of such distances are smallest when x sits among the points.
  • |x−a| = distance of x from a on the number line.
  • |x−p|+|x−q| is minimised for any x between p and q; minimum value = |p−q|.
  • |x−p| = |x−q| at the midpoint x = (p+q)/2.
  • e.g. |x−2| + |x−7| ≥ 5, achieved for any x in [2, 7].
29Sum of squares identity trick
  • Plain English: squares can't be negative, so if a bunch of squares add to 0 every single one must be 0.
  • If a sum of squares equals 0, each square = 0: e.g. (x−2y)² + (y−z)² = 0 ⇒ x = 2y and y = z.
  • Group given expressions into perfect squares to pin exact values.
  • e.g. (a−3)² + (b+1)² = 0 forces a = 3 and b = −1.
30Cauchy / vector identity
  • Plain English: this identity links two "sum-of-squares" products to two cross-terms, handy when three of the four pieces are given.
  • (a²+b²)(x²+y²) = (ax+by)² + (ay−bx)².
  • Useful when given a²+b², x²+y² and ax+by to find ay−bx.
  • e.g. (1²+2²)(3²+4²) = 5·25 = 125 = 11² + 2² = (1·3+2·4)² + (1·4−2·3)².
49 CAT questions

Progressions & Series, CAT PYQs

Progressions & Series

HardCAT 2003

The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, f, f, f, f, f… is

  • (1) u
  • (2) v
  • (3) w
  • (4) x
Show solution
(4) x. The nth letter of the alphabet is written n times, so the last position of the nth letter is 1 + 2 + … + n = n(n+1)/2. For n = 23, this is 23·24/2 = 276; for n = 24 it is 24·25/2 = 300. So positions 277-300 all carry the 24th letter, which is "x". Hence the 288th term is x.
HardCAT 2003

There are 8436 steel balls, each with a radius of 1 centimetre, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

  • (1) 34
  • (2) 38
  • (3) 36
  • (4) 32
Show solution
(3) 36. Each layer holds a triangular number k(k+1)/2 of balls; the cumulative total of the first n triangular numbers is n(n+1)(n+2)/6. Setting n(n+1)(n+2)/6 = 8436 gives n = 36.
HardCAT 2003

The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of the 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

  • (1) 1st
  • (2) 9th
  • (3) 12th
  • (4) None of these
Show solution
(3) 12th. T₃ + T₁₅ = T₆ + T₁₁ + T₁₃ ⇒ (a+2d) + (a+14d) = (a+5d) + (a+10d) + (a+12d) ⇒ 2a + 16d = 3a + 27d ⇒ a + 11d = 0. Since T₁₂ = a + 11d, the 12th term of the AP is 0.
HardCAT 2003

Let T be the set of integers {3, 11, 19, 27, … 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

  • (1) 32
  • (2) 28
  • (3) 29
  • (4) 30
Show solution
(4) 30. T is an AP with first term 3, common difference 8 and last term 467, so it has 59 terms. Pairs adding to 470 are (3, 467), (11, 459), …, 29 such pairs, and the lone middle term 235 (which has no partner). Picking at most one from each pair gives 29 elements, plus 235, so the maximum is 30.
ModerateCAT 2003

Let S = 2x + 5x² + 9x³ + 14x⁴ + 20x⁵ ……infinity. The coefficient of nth term is n(n+3)/2. The sum S is:

  • (1) x(2−x)/(1−x)³
  • (2) (2−x)/(1−x)³
  • (3) x(2−x)/(1−x)²
  • (4) None of these
Show solution
(1) x(2−x)/(1−x)³. Writing the nth coefficient as n(n+3)/2 = ½(n² + 3n) and using Σ nxⁿ = x/(1−x)² and Σ n²xⁿ = x(1+x)/(1−x)³, the sum simplifies to x(2−x)/(1−x)³.
ModerateCAT 2004

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

  • (1) 0
  • (2) −1
  • (3) 1
  • (4) Not unique
Show solution
(1) 0. If Sₚ = S_q (p ≠ q) for an AP, then S_(p+q) = 0. Here S₁₁ = S₁₉, so S₃₀ = S_(11+19) = 0.
HardCAT 2005

If a₁ = 1 and aₙ₊₁ − 3aₙ + 2 = 4n for every positive integer n, then a₁₀₀ equals

  • (1) 3⁹⁹ − 200
  • (2) 3⁹⁹ + 200
  • (3) 3¹⁰⁰ − 200
  • (4) 3¹⁰⁰ + 200
Show solution
(3) 3¹⁰⁰ − 200. aₙ₊₁ = 3aₙ + 4n − 2. Computing a₂ = 5 = 3²−4, a₃ = 21 = 3³−6, a₄ = 73 = 3⁴−8 ⇒ aₙ = 3ⁿ − 2n. So a₁₀₀ = 3¹⁰⁰ − 200.
ModerateCAT 2003

The infinite sum 1 + 4/7 + 9/7² + 16/7³ + 25/7⁴ + …. equals:

  • (1) 27/14
  • (2) 21/13
  • (3) 49/27
  • (4) 256/147
Show solution
(3) 49/27. Let S = 1 + 4/7 + 9/7² + 16/7³ + …. Then S/7 = 1/7 + 4/7² + 9/7³ + …; subtracting gives 6S/7 = 1 + 3/7 + 5/7² + 7/7³ + …. Repeating the divide-and-subtract step: 6S/49 = 1 + 2(1/7 + 1/7² + …) = 1 + 2·(1/7)/(1−1/7) = 4/3. Hence S = 49·(4/3)/36 = 49/27.
ModerateCAT 2006

Let S₁ be a square of side a. Another square S₂ is formed by joining the mid-points of the sides of S₁. The same process is applied to S₂ to form yet another square S₃, and so on. If A₁, A₂, A₃, …… be the areas and P₁, P₂, P₃, …… be the perimeters of S₁, S₂, S₃, ……, respectively, then the ratio (P₁ + P₂ + P₃ + ……)/(A₁ + A₂ + A₃ + ……) equals:

  • (1) 2(1+√2)/a
  • (2) 2(2−√2)/a
  • (3) 2(2+√2)/a
  • (4) 2(1+2√2)/a
Show solution
(3) 2(2+√2)/a. Each new square has side 1/√2 times the previous, so perimeters form a GP with ratio 1/√2 and areas a GP with ratio 1/2. ΣP = 4a/(1 − 1/√2) and ΣA = a²/(1 − 1/2) = 2a². Dividing and rationalising gives 2(2+√2)/a.
HardCAT 2007

A function f satisfies f(1) = 3600 and f(1) + f(2) + … + f(n) = n²·f(n) for all n > 1. What is f(9)?

  • (1) 80
  • (2) 240
  • (3) 200
  • (4) 100
Show solution
(1) 80. Subtracting consecutive relations gives (n²−1)f(n) = (n−1)²f(n−1) ⇒ f(n) = ((n−1)/(n+1))·f(n−1). Telescoping from f(1) = 3600: f(9) = 3600·(2/(9·10)) = 80.
HardCAT 2008

The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is:

  • (1) 78
  • (2) 19
  • (3) 20
  • (4) 77
Show solution
(3) 20. The first common term is 21; the common terms form an AP with common difference LCM(4, 5) = 20, i.e. 21, 41, 61, …. The largest such term not exceeding the smaller cap 417 gives 20 common terms.
HardCAT 2008

Consider the set S = {1, 2, 3, …., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and with 1000 and have at least 3 elements?

  • (1) 3
  • (2) 4
  • (3) 6
  • (4) 7
Show solution
(4) 7. The common difference d must divide 1000 − 1 = 999 = 3³·37 and give at least 3 terms, i.e. (999/d) + 1 ≥ 3 ⇒ d ≤ 499.5. Divisors of 999 not exceeding 499 are 1, 3, 9, 27, 37, 111, 333, that is 7 progressions.
HardCAT 2008

Find the sum √(1 + 1/1² + 1/2²) + √(1 + 1/2² + 1/3²) + … + √(1 + 1/2007² + 1/2008²).

  • (1) 2008 − 1/2008
  • (2) 2007 − 1/2007
  • (3) 2007 − 1/2008
  • (4) 2008 − 1/2007
Show solution
(1) 2008 − 1/2008. Using the identity √(1 + 1/n² + 1/(n+1)²) = 1 + 1/n − 1/(n+1), the sum telescopes. With n running from 1 to 2007, total = 2007 + (1/1 − 1/2008) = 2008 − 1/2008.
ModerateCAT 2017

If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

  • (1) 2 : 3
  • (2) 3 : 2
  • (3) 3 : 4
  • (4) 4 : 3
Show solution
(1) 2 : 3. (a+6d)² = (a+2d)(a+16d) ⇒ a² + 12ad + 36d² = a² + 18ad + 32d² ⇒ 4d² = 6ad ⇒ a : d = 2 : 3.
ModerateCAT 2017

Let a₁, a₂, …, aₙ, be an arithmetic progression with a₁ = 3 and a₂ = 7. If a₁ + a₂ + … + aₙ = 1830, then what is the smallest positive integer m such that m(a₁ + a₂ + … + aₙ) > 1830?

  • (1) 8
  • (2) 9
  • (3) 10
  • (4) 11
Show solution
(2) 9. Here aₙ = 4n − 1, so the sum of the first n terms is Sₙ = n(2n + 1); Sₙ = 1830 gives n = 30. Working with the partial sums, the smallest positive integer m satisfying the stated inequality is 9.
ModerateCAT 2017TITA

Let a₁, a₂, a₃, a₄, a₅ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a₃. If the sum of the numbers in the new sequence is 450, then a₅ is

Show solution
51. The new even sequence is 2a₃ − 8, 2a₃ − 6, 2a₃ − 4, 2a₃ − 2, 2a₃; its sum is 10a₃ − 20 = 450 ⇒ a₃ = 47. The odd numbers step by 2, so a₅ = a₃ + 4 = 51.
HardCAT 2017

An infinite geometric progression a₁, a₂, a₃, … has the property that aₙ = 3(aₙ₊₁ + aₙ₊₂ + …) for every n ≥ 1. If the sum a₁ + a₂ + a₃ + … = 32, then a₅ is

  • (1) 1/32
  • (2) 2/32
  • (3) 3/32
  • (4) 4/32
Show solution
(3) 3/32. aₙ = 3·aₙ₊₁/(1 − r) gives 1 = 3r/(1 − r) ⇒ r = 1/4. With sum 32 = a/(1 − r) = a/(3/4), a = 24. So a₅ = 24·(1/4)⁴ = 24/256 = 3/32.
HardCAT 2017

If a₁ = 1/(2×5), a₂ = 1/(5×8), a₃ = 1/(8×11), …, then a₁ + a₂ + … + a₁₀₀ is

  • (1) 25/151
  • (2) 1/2
  • (3) 1/4
  • (4) 111/55
Show solution
(1) 25/151. Each term telescopes: aₖ = (1/3)[1/(3k − 1) − 1/(3k + 2)]. Summing k = 1 to 100 = (1/3)[1/2 − 1/302] = (1/3)(150/302) = 25/151.
HardCAT 2006

Consider a sequence where the nth term, tₙ = n/(n+2), n = 1, 2, … . The value of t₃ × t₄ × t₅ × … × t₅₃ equals:

  • (1) 2/495
  • (2) 2/477
  • (3) 12/55
  • (4) 1/1485
Show solution
(1) 2/495. Product = (3·4·…·53)/(5·6·…·55), which telescopes to (3·4)/(54·55) = 12/2970 = 2/495.
ModerateCAT 2017TITA

If a₁ + a₂ + a₃ + … + aₙ = 3(2ⁿ⁺¹ − 2), for every n ≥ 1, then a₁₁ equals

Show solution
6144. aₙ = Sₙ − Sₙ₋₁ = 3(2ⁿ⁺¹ − 2ⁿ) = 3·2ⁿ. So a₁₁ = 3·2¹¹ = 3·2048 = 6144.
ModerateCAT 2017

Let a₁, a₂, … be integers such that a₁ − a₂ + a₃ − a₄ + … + (−1)ⁿ⁻¹ aₙ = n, for all n ≥ 1. Then a₅₁ + a₅₂ + … + a₁₀₂₃ equals

  • (1) 0
  • (2) 1
  • (3) 10
  • (4) −1
Show solution
(2) 1. Subtracting consecutive defining equations gives aₙ + aₙ₋₁ = ±1 in a fixed pattern; pairing the terms from a₅₁ through a₁₀₂₃ cancels in pairs, leaving a total of 1.
HardCAT 2018

Given an equilateral triangle T₁ with side 24 cm, a second triangle T₂ is formed by joining the midpoints of the sides of T₁. Then a third triangle T₃ is formed by joining the midpoints of the sides of T₂. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T₁, T₂, T₃, … will be

  • (1) 188√3
  • (2) 248√3
  • (3) 164√3
  • (4) 192√3
Show solution
(4) 192√3. Each new triangle has half the side, so its area is one-quarter of the previous: areas form a GP with ratio 1/4. Area of T₁ = (√3/4)·24² = 144√3. Sum = 144√3/(1 − 1/4) = 144√3·(4/3) = 192√3.
HardCAT 2018

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in arithmetic progression then the common ratio of the geometric progression is

  • (1) 3/6
  • (2) 1/6
  • (3) 5/2
  • (4) 3/2
Show solution
(3) 5/2. Let x = a, y = ar, z = ar². The AP condition gives 2(16y) = 5x + 12z ⇒ 32r = 5 + 12r² ⇒ 12r² − 32r + 5 = 0 ⇒ r = 5/2 or 1/6. Since x < y < z needs r > 1, r = 5/2.
HardCAT 2018

Let a₁, a₂…a₅₂ be positive integers such that a₁ < a₂ < … < a₅₂. Suppose, their arithmetic mean is one less than arithmetic mean of a₂, a₃, …. a₅₂. If a₅₂ = 100, then the largest possible value of a₁ is

  • (1) 48
  • (2) 20
  • (3) 23
  • (4) 45
Show solution
(3) 23. Let A be the mean of a₂, …, a₅₂ (51 terms, sum 51A). The mean of all 52 terms is A − 1, so (a₁ + 51A)/52 = A − 1 ⇒ a₁ = A − 52. To maximise a₁, maximise A: with 51 distinct integers ≤ 100 ending at a₅₂ = 100, the largest mean is for 50, 51, …, 100, giving A = 75. Hence a₁ = 75 − 52 = 23.
ModerateCAT 2018TITA

The value of the sum 7 × 11 + 11 × 15 + 15 × 19 + … + 95 × 99 is

Show solution
80707. The nth term is (4n + 3)(4n + 7) = 16n² + 40n + 21, with 23 terms (the first factors 7, 11, …, 95). Σ over n = 1 to 23 = 16·Σn² + 40·Σn + 21·23 = 80707.
HardCAT 2018TITA

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = n(n+1)/2 and v = (y+z)/2. If x ≥ z, then the minimum possible value of x is

Show solution
105. x + y + z = 240 and x + y + z + u + v = 375, so u + v = 135. With u = (x+y)/2 and v = (y+z)/2, (x + 2y + z)/2 = 135 ⇒ x + 2y + z = 270 ⇒ y = 30 and x + z = 210. With x ≥ z, the minimum value of x is 105.
ModerateCAT 2018TITA

If (2n + 1) + (2n + 3) + (2n + 5) + … + (2n + 47) = 5280, then what is the value of 1 + 2 + 3 + …. + n?

Show solution
4851. There are 24 terms; their sum = 24·(2n) + (1 + 3 + … + 47) = 48n + 576 = 5280 ⇒ n = 98. Then 1 + 2 + … + 98 = 98·99/2 = 4851.
ModerateCAT 2018TITA

Let t₁, t₂,… be real with t₁ + t₂ + … + tₙ = 2n² + 9n + 13 for every integer n ≥ 2. If t_k = 103, then k equals

Show solution
24. tₙ = Sₙ − Sₙ₋₁ = (2n²+9n+13) − (2(n−1)²+9(n−1)+13) = 4n + 7. Set 4k + 7 = 103 ⇒ k = 24.
HardCAT 2019

The number of common terms in the two sequences: 15, 19, 23, 27, …., 415 and 14, 19, 24, 29, …, 464 is:

  • (1) 21
  • (2) 20
  • (3) 18
  • (4) 19
Show solution
(4) 19. The first common term is 19; the common terms form an AP with common difference LCM(4, 5) = 20: 19, 39, 59, …. Counting up to the smaller cap 415 gives 19 common terms.
HardCAT 2019

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

  • (1) (1003)¹⁵ + 6
  • (2) (997)¹⁵ − 3
  • (3) (997)2¹⁴ + 3
  • (4) (1003)2¹⁵ − 3
Show solution
(4) (1003)2¹⁵ − 3. Since pₙ₊₁ + 3 = 2(pₙ + 3), the quantity (pₙ + 3) doubles each year. From 2019 to 2034 is 15 years, so p + 3 = (1000 + 3)·2¹⁵ ⇒ p = (1003)2¹⁵ − 3.
ModerateCAT 2020

If x₁ = −1 and xₘ = xₘ₊₁ + (m + 1) for every positive integer m, then x₁₀₀ equals

  • (1) −5151
  • (2) −5150
  • (3) −5051
  • (4) −5050
Show solution
(4) −5050. xₘ₊₁ = xₘ − (m+1): x₂ = −3, x₃ = −6, x₄ = −10… ⇒ xₙ = −n(n+1)/2. x₁₀₀ = −100·101/2 = −5050.
ModerateCAT 2020

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by:

  • (1) 3
  • (2) 6
  • (3) 5
  • (4) 4
Show solution
(4) 4. The lowest 9 scores sum to 9·42 = 378; the highest 9 sum to 9·47 = 423. The total of all 10 ranges between 378 + (highest, ≥ 47) and 423 + (lowest, ≤ 42); bounding the extra score gives a total-mean range of width 4.
HardCAT 2020

Let the mth and nth terms of a geometric progression be 3/4 and 12, respectively, where m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n − m is:

  • (1) 6
  • (2) −4
  • (3) −2
  • (4) 2
Show solution
(3) −2. 12 ÷ (3/4) = 16 = r^(n−m). For integer r: r = 2 ⇒ n − m = 4 ⇒ sum 6; r = −2 ⇒ n − m = 4 ⇒ sum 2; r = 4 ⇒ n − m = 2 ⇒ sum 6; r = −4 ⇒ n − m = 2 ⇒ sum −2. The smallest is −2.
HardCAT 2021 · Slot 1

If x₀ = 1, x₁ = 2, and xₙ₊₂ = (1 + xₙ₊₁)/xₙ, n = 0, 1, 2, 3, …, then x₂₀₂₁ is equal to

  • (1) 1
  • (2) 2
  • (3) 3
  • (4) 4
Show solution
(2) 2. The sequence is periodic with period 5: x₀, x₁, x₂, x₃, x₄ = 1, 2, 3, 2, 1, then it repeats. Since 2021 mod 5 = 1, x₂₀₂₁ = x₁ = 2.
HardCAT 2021 · Slot 2

Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals

  • (1) 8
  • (2) 12
  • (3) 10
  • (4) 14
Show solution
(4) 14. In an AP, x + y + z = 3y, so xyz = 5·3y = 15y ⇒ xz = 15. The factor pairs of 15 with y the average and y − x > 2 give x = 1, z = 15 (y = 8), so z − x = 14.
ModerateCAT 2021 · Slot 2

For a sequence of real numbers x₁, x₂, …, xₙ, if x₁ − x₂ + x₃ − … + (−1)ⁿ⁺¹ xₙ = n² + 2n for all natural numbers n, then the sum x₄₉ + x₅₀ equals

  • (1) −2
  • (2) 2
  • (3) −200
  • (4) 200
Show solution
(1) −2. Adding the equations for n = 49 and n = 50 (which have opposite signs on x₅₀) isolates x₄₉ + x₅₀; it works out to −2.
ModerateCAT 2021 · Slot 3

Consider a sequence of real numbers x₁, x₂, x₃, … such that xₙ₊₁ = xₙ + n − 1 for all n ≥ 1. If x₁ = −1, then x₁₀₀ is equal to

  • (1) 4949
  • (2) 4850
  • (3) 4849
  • (4) 4950
Show solution
(2) 4850. x₁₀₀ = x₁ + Σ(n − 1) for n = 1 to 99 = −1 + (0 + 1 + … + 98) = −1 + 98·99/2 = −1 + 4851 = 4850.
ModerateCAT 2022 · Slot 1

For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n + 2n²). If the nth term of the progression is divisible by 9, then the smallest possible value of n is:

  • (1) 8
  • (2) 7
  • (3) 4
  • (4) 9
Show solution
(2) 7. aₙ = Sₙ − Sₙ₋₁ = (n + 2n²) − ((n−1) + 2(n−1)²) = 4n − 1. For 9 | (4n − 1): 4n ≡ 1 (mod 9) ⇒ n ≡ 7 (mod 9). The smallest such n is 7.
HardCAT 2022 · Slot 2TITA

The average of a non-decreasing sequence of N numbers a₁, a₂, ……, aN is 300. If a₁ is replaced by 6a₁, the new average becomes 400. Then, the number of possible values of a₁ is:

Show solution
14. The total rises by 5a₁ = 100N, so a₁ = 20N. As a₁ is the smallest of N non-decreasing terms with average 300, N can range from 2 to 15, giving 14 possible values.
ModerateCAT 2022 · Slot 2

On day one, there are 100 particles in a laboratory experiment. On day n, where n ≥ 2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals:

  • (1) 19
  • (2) 17
  • (3) 16
  • (4) 18
Show solution
(1) 19. On day n the count is multiplied by (n+1)/n, so after the telescoping product the count on day m is 100·(m+1)/2. Setting 100·(m+1)/2 = 1000 ⇒ m + 1 = 20 ⇒ m = 19.
HardCAT 2022 · Slot 2

Consider the arithmetic progression 3, 7, 11, … and let Aₙ denote the sum of the first n terms of this progression. Then the value of (1/25)·Σ(n=1 to 25) Aₙ is:

  • (1) 442
  • (2) 404
  • (3) 455
  • (4) 415
Show solution
(3) 455. Aₙ = (n/2)(2·3 + (n−1)·4) = 2n² + n. Σ over n = 1 to 25 = 2·Σn² + Σn = 2·5525 + 325 = 11375. Dividing by 25 gives 455.
ModerateCAT 2022 · Slot 3

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

  • (1) 2442
  • (2) 3333
  • (3) 2592
  • (4) 2222
Show solution
(4) 2222. Distinct arrangements = 4!/2! = 12 (the digit 1 repeats). Each digit occupies each place 3 times, so the digit-sum per place = 3·(1 + 4 + 2 + 1) = 24. Total of all numbers = 24·1111 = 26664; mean = 26664/12 = 2222.
ModerateCAT 2022 · Slot 3TITA

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value of n is

Show solution
19. Since aₙ + 3 doubles each day, aₙ + 3 = 5·2ⁿ⁻¹, i.e. aₙ = 5·2ⁿ⁻¹ − 3. Requiring 5·2ⁿ⁻¹ − 3 > 1,000,000 ⇒ 2ⁿ⁻¹ > 200,000.6 ⇒ n − 1 ≥ 18 ⇒ n = 19.
ModerateCAT 2023 · Slot 1TITA

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Show solution
92. Total = 25·50 = 1250. To maximise the toppers' score, minimise the other 20: distinct integers from 30 upward, i.e. 30, 31, …, 49, summing to 790. The remaining 1250 − 790 = 460 is shared by 5 toppers ⇒ 92 each.
HardCAT 2023 · Slot 2

Let both the series a₁, a₂, a₃, … and b₁, b₂, b₃, … be in arithmetic progression such that the common differences of both the series are prime numbers. If a₅ = b₉, a₁₉ = b₁₉ and b₂ = 0, then a₁₁ equals

  • (1) 86
  • (2) 84
  • (3) 79
  • (4) 83
Show solution
(3) 79. Let the common differences be primes p (for aₙ) and q (for bₙ). a₁₉ = b₁₉ and a₅ = b₉ give relations that, with b₂ = 0, force p = 7 and q = 5. Then a₁₁ works out to 79.
ModerateCAT 2023 · Slot 2TITA

Let aₙ and bₙ be two sequences such that aₙ = 13 + 6(n−1) and bₙ = 15 + 7(n−1) for all natural numbers n. Then, the largest three-digit integer that is common to both these sequences, is

Show solution
967. The aₙ values are 13, 19, 25, … (step 6); the bₙ values are 15, 22, 29, … (step 7). Common terms step by LCM(6, 7) = 42 from the first common value 43; the largest three-digit such term is 967.
HardCAT 2023 · Slot 3

Let aₙ = 46 + 8n and bₙ = 98 + 4n be two sequences for natural numbers n ≤ 100. Then, the sum of all terms common to both the sequences is

  • (1) 14602
  • (2) 14798
  • (3) 15000
  • (4) 14900
Show solution
(4) 14900. aₙ runs 54, 62, …, 846 (step 8); bₙ runs 102, 106, …, 498 (step 4). Common terms form an AP with common difference LCM(8, 4) = 8 within the overlapping range; summing them gives 14900.

CAT 2024 & 2025, recent

ModerateCAT 2024 · Slot 1

Suppose x₁, x₂, x₃, …, x₁₀₀ are in arithmetic progression such that x₅ = −4 and 2x₆ + 2x₉ = x₁₁ + x₁₃. Then, x₁₀₀ equals

  • (A) −194
  • (B) −196
  • (C) 204
  • (D) 206
Show solution
(A) −194. With xₙ = a + (n−1)d: 2x₆ + 2x₉ = x₁₁ + x₁₃ ⇒ 2(a+5d) + 2(a+8d) = (a+10d) + (a+12d) ⇒ 4a + 26d = 2a + 22d ⇒ a = −2d. Then x₅ = a + 4d = −2d + 4d = 2d = −4 ⇒ d = −2, a = 4. So x₁₀₀ = a + 99d = 4 + 99(−2) = −194.
HardCAT 2025 · Slot 2

Let aₙ be the nᵗʰ term of a decreasing infinite geometric progression. If a₁ + a₂ + a₃ = 52 and a₁a₂ + a₂a₃ + a₃a₁ = 624, then the sum of this geometric progression is

  • (A) 57
  • (B) 54
  • (C) 60
  • (D) 63
Show solution
(B) 54. With a₁ = a, ratio r: a(1 + r + r²) = 52 and a²r(1 + r + r²) = 624 ⇒ ar = 624/52 = 12 (dividing). Then a²r·(1+r+r²) = 624 with a(1+r+r²) = 52 gives ar = 12; combined with a(1+r+r²)=52 and a₂ = ar = 12 ⇒ a + 12 + 12r = 52 ⇒ a + 12r = 40, and a·r = 12 ⇒ a = 36, r = 1/3 (decreasing GP needs |r| < 1). Sum = a/(1 − r) = 36/(1 − 1/3) = 54.