Progression CAT PYQs , 56 Questions with Solutions

Master Progression for CAT 2026 with practice questions and detailed explanations

CAT 2025

Progressions (Arithmetic, Geometric progressions, series sums) are tested occasionally. They fall under algebra/series. CAT 2017 did not highlight series, but CAT 2018 had at least one series question (e.g. sum of a sequence, counted in the 3 “Modern Maths” Qs) . CAT 2019’s analysis doesn’t mention it explicitly, implying maybe none or trivial presence that year. CAT 2020 slot2 had 1 progression question , slot1 and slot3 had 0. CAT 2021 had a few: about 1–2 questions on series across slots . CAT 2022 also included progression in one slot (slot2 had 3, others 1) . By CAT 2023–2024, progressions remained minor but present (roughly 1 question per paper) . So, progressions aren’t heavily featured, but CAT often throws in one question involving an arithmetic or geometric series sum or nth term.

Weightage Over Past Years

YearQ.NODifficulty Level
20253Hard
20245Hard
202310Hard
20226Medium
20216Medium
20206Medium
20199Medium
20185Medium
20176Hard

CAT 2025 Progression questions

Question 1

Slot-1

In the set of consecutive odd numbers {1,3,5,…,57}\{1, 3, 5, \ldots, 57\}, there is a number kk such that the sum of all elements less than kk is equal to the sum of all elements greater than kk. Then, kk equals

41

39

37

43

Show Solution

Question 2

Slot-2

Let ana _ { n } be the nth n ^ { \text {th } } term of a decreasing infinite geometric progression. If a1+a2+a3=52a _ { 1 } + a _ { 2 } + a _ { 3 } = 52 and a1a2+a2a3+a3a1=624a _ { 1 } a _ { 2 } + a _ { 2 } a _ { 3 } + a _ { 3 } a _ { 1 } = 624, then the sum of this geometric progression is

6363

6060

5454

5757

Show Solution

Question 3

Slot-3

In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is

Show Solution

CAT 2024 Progression questions

Question 1

Slot-1

Suppose x1,x2,x3,…,x100x_1, x_2, x_3, \ldots, x_{100} are in arithmetic progression such that x5=−4x_5 = -4 and 2×x6+2×x9=x11+x132 \times x_6 + 2 \times x_9 = x_{11} + x_{13}. Then, x100x_{100} equals

204

-194

-196

206

Show Solution

Question 2

Slot-2

The sum of the infinite series 15(15−17)+(15)2((15)2−(17)2)+(15)3((15)3−(17)3)+⋯\frac{1}{5}\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{5}\right)^2\left(\left(\frac{1}{5}\right)^2-\left(\frac{1}{7}\right)^2\right)+\left(\frac{1}{5}\right)^3\left(\left(\frac{1}{5}\right)^3-\left(\frac{1}{7}\right)^3\right)+\cdots is equal to

5408\frac{5}{408}

7816\frac{7}{816}

7408\frac{7}{408}

5816\frac{5}{816}

Show Solution

Question 3

Slot-3

Consider the sequence t1=1t_1 = 1, t2=−1t_2 = -1, and

tn=(n−3n−1)tn−2for n≥3.t_n = \left( \frac{n - 3}{n - 1} \right)t_{n-2} \quad \text{for } n \geq 3.

Then, the value of the sum

1t2+1t4+1t6+⋯+1t2024\frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \cdots + \frac{1}{t_{2024}}

is:

-1023132

-1022121

-1024144

-1026169

Show Solution

Question 4

Slot-1

Based on the given pattern find the next term of the given series: 5, 6, 16, 57, 244, 1245, ?

7506

None of these

5500

4047

Show Solution

Question 5

Slot-1

In a school, the number of students in each class, from Class I to X, in that order, are in an arithmetic progression. The total number of students from Class I to V is twice the total number of students from Class VI to X. If the total number of students from Class I to IV is 462, how many students are there in Class VI?

93

88

83

77

None of the remaining options is correct

Show Solution

CAT 2023 Progression questions

Question 1

Slot-1

For some positive and distinct real numbers x,yx, y and zz , if 1y+z\frac{1}{\sqrt{y}+\sqrt{z}} is the arithmetic mean of 1x+z\frac{1}{\sqrt{x}+\sqrt{z}} and 1x+y\frac{1}{\sqrt{x}+\sqrt{y}} , then the relationship which will always hold true, is

x,y\sqrt{x}, \sqrt{y} and z\sqrt{z} are in arithmetic progression

x,z\sqrt{x}, \sqrt{z} and y\sqrt{y} are in arithmetic progression

y,xy, x and zz are in arithmetic progression

x,yx, y and zz are in arithmetic progression

Show Solution

Question 2

Slot-1

Separately, Jack and Sristi invested the same amount of money in a stock market. Jack’s invested amount kept getting reduced by 50%50\% every month. Sristi’s investment also reduced every month, but in an arithmetic progression with a common difference of Rs. 1500015000. They both withdrew their respective amounts at the end of the sixth month. They observed that if they had withdrawn their respective amounts at the end of the fourth month, the ratio of their amounts would have been the same as the ratio after the sixth month. What amount of money was invested by Jack in the stock market?

Rs. 100000100000

Rs. 120000120000

Rs. 150000150000

Rs. 180000180000

None of the above

Show Solution

Question 3

Slot-1

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 n^{\text {th }} day exceeds one million, then the lowest possible value of nn is

Show Solution

Question 4

Slot-2

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

Show Solution

Question 5

Slot-2

Let both the series a1,a2,a3,…a_1, a_2, a_3, \ldots and b1,b2,b3…b_1, b_2, b_3 \ldots be in arithmetic progression such that the common differences of both the series are prime numbers. If a5=b9,a19=b19a_5=b_9, a_{19}=b_{19} and b2=0b_2=0 , then a11a_{11} equals

79

83

86

84

Show Solution

Question 6

Slot-3

The value of 1+(1+13)14+(1+13+19)116+(1+13+19+127)164+⋯1 + \left(1 + \frac{1}{3}\right) \frac{1}{4} + \left(1 + \frac{1}{3} + \frac{1}{9}\right) \frac{1}{16} + \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}\right) \frac{1}{64} + \cdots, is

1513\frac{15}{13}

1611\frac{16}{11}

2712\frac{27}{12}

158\frac{15}{8}

Show Solution

Question 7

Slot-3

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

15000

14900

14602

14798

Show Solution

Question 8

Slot-1

Consider the sequence an+1=11+1ana_{n+1} = \dfrac{1}{1 + \dfrac{1}{a_n}} for n=1,2,…,2008,2009n = 1, 2, \ldots, 2008, 2009 where a1=1a_1 = 1. Find the value of a1a2+a2a3+a3a4+…+a2008a2009a_1a_2 + a_2a_3 + a_3a_4 + \ldots + a_{2008}a_{2009}.

20091000\dfrac{2009}{1000}

20092008\dfrac{2009}{2008}

20082009\dfrac{2008}{2009}

60002009\dfrac{6000}{2009}

20086000\dfrac{2008}{6000}

Show Solution

Question 9

Slot-1

There are three sections in a question paper and each section has 10 questions. First section only has multiple-choice questions, and 2 marks will be awarded for each correct answer. For each wrong answer, 0.50.5 marks will be deducted. Any unattempted question in this section will be treated as a wrong answer. Each question in the second section carries 3 marks, whereas each question in the third section carries 5 marks. For any wrong answer or un-attempted question in the second and third sections, no marks will be deducted. A student’s score is the addition of marks obtained in all the three sections. What is the sixth highest possible score?

92.5

94

95.5

95

None of the above

Show Solution

Question 10

Slot-1

If in a decreasing AP, sum of all the terms except the first term is −75 while the sum of all the terms is 0. If the difference between the 9th term and the 13th term is 20, the find the first term of the AP?

25

30

35

None of these

Show Solution

CAT 2022 Progression questions

Question 1

Slot-1

For any natural number nn , suppose the sum of the first nn terms of an arithmetic progression is (n+2n2)\left(n+2 n^2\right) . If the nth n^{\text {th }} term of the progression is divisible by 9 , then the smallest possible value of nn is

4

8

7

9

Show Solution

Question 2

Slot-1

A marble is dropped from a height of 3 metres onto the ground. After hitting the ground, it bounces and reaches 80% of the height from which it was dropped. This repeats multiple times. Each time it bounces, the marble reaches 80% of the height previously reached. Eventually, the marble comes to rest on the ground. What is the maximum distance that the marble travels from the time it was dropped until it comes to rest?

15 m

27 m

24 m

12 m

30 m

Show Solution

Question 3

Slot-1

If both the sequences x,a1,a2,yx, a_1, a_2, y and x,b1,b2,zx, b_1, b_2, z are in A.P. and it is given that y>zy > z and z<xz < x, then which of the following values can (a1−a2)(b1−b2)\frac{(a_1 - a_2)}{(b_1 - b_2)} possibly take?

2

5

-3

1

0

Show Solution

Question 4

Slot-2

Consider the arithmetic progression 3,7,11,…3,7,11, \ldots and let AnA_n denote the sum of the first nn terms of this progression. Then the value of 125∑n=125An\frac{1}{25} \sum_{n=1}^{25} A_n is

404

442

455

415

Show Solution

Question 5

Slot-2

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

19

16

17

18

Show Solution

Question 6

Slot-3

The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is

Show Solution

CAT 2021 Progression questions

Question 1

Slot-1

Given below are two statements Statement I: In the sequence of numbers 30, 90, 182, 306, 462, P….. the term P is 650. Statement II: There are 8 digits in 9⁸ when it is expressed in decimal form. In light of the above statements, choose the correct answer from the options given below

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Show Solution

Question 2

Slot-1

Natural numbers are divided into groups as follows:

  • Group 1: (1)
  • Group 2: (2, 3, 4)
  • Group 3: (5, 6, 7, 8, 9)
  • Group 4: (10 … 16)
    …and so on, where the k‑th group contains (2k–1) numbers.
    What is the sum of the numbers in the 15th group?

6119

7471

4941

6090

Show Solution

Question 3

Slot-2

For a sequence of real numbers x1,x2,…,xnx_1, x_2, \ldots, x_n, if x1−x2+x3−⋯+(−1) n+1xn=n2+2nx_1 - x_2 + x_3 - \cdots + (-1)^{\,n+1} x_n = n^2 + 2n for all natural numbers nn, then the sum x49+x50x_{49} + x_{50} equals

2

-2

200

-200

Show Solution

Question 4

Slot-2

Find the sum of the following series (with infinite terms): 2√2, 4/√3, 4√2/3, ……

2√3(√3 + √2)

2√3(√3 − √2)

2√6(√3 + √2)

2√6(√3 − √2)

Show Solution

Question 5

Slot-2

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

8

10

14

12

Show Solution

Question 6

Slot-3

Consider a sequence of real numbers x1,x2,x3,…x_{1}, x_{2}, x_{3}, \ldots such that xn+1=xn+n−1x_{n+1}=x_{n}+n-1 for all n≥1.n \geq 1 . If x1=−1x_{1}=-1 then x100x_{100} is equal to

4949

4849

4850

4950

Show Solution

CAT 2020 Progression questions

Question 1

Slot-2

Let the m-th and n-th terms of a Geometric progression be 34\frac{3}{4} and 12, respectively, when m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is

-4

-2

6

2

Show Solution

Question 2

Slot-1

Find the missing number in the series:

1680, 1168, 1119, 903, ?, 814

None of these

864

814

878

858

Show Solution

Question 3

Slot-1

Find the missing number in the series:

110, 55, 55, 110, ?, 3520

425

380

440

None of these

450

Show Solution

Question 4

Slot-1

Find the missing number in the series:

8, 5, 7, 18, 80, ?

656

None of these

126

642

428

Show Solution

Question 5

Slot-1

Find the missing number in the series:

11, 12, 17, 30, 59, ?

None of these

125

120

100

110

Show Solution

Question 6

Slot-1

Find the missing number in the series:

?, 237, 366, 473, 547, 577

101

105

78

97

None of these

Show Solution

CAT 2019 Progression questions

Question 1

Slot-1

If the population of a town is pp in the beginning of any year, then it becomes 3+2p3 + 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

1003×315−61003 \times 3^{15} - 6

977×425−5977\times 4^{25} - 5

1003×215−31003 \times 2^{15} - 3

(977)214+3(977)^{214} + 3

Show Solution

Question 2

Slot-1

If a1+a2+a3+⋯+an=3(2n+1−2),a_1 + a_2 + a_3 + \dots + a_n = 3(2^{n+1} - 2), then find the value of a11. a_{11}.

Show Solution

Question 3

Slot-1

If a1,a2,…a_1, a_2, \ldots are in Arithmetic Progression (A.P.), then the value of the expression

1a1+a2+1a2+a3+⋯+1an+an+1\dfrac{1}{\sqrt{a_1} + \sqrt{a_2}} + \dfrac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \dfrac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}
is equal to:

n / (√a₁ + √aₙ₊₁)

n−1a1+an\frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}}

na1−an+1\frac{n}{\sqrt{a_1} - \sqrt{a_{n+1}}}

n−1a1+an−1\frac{n - 1}{\sqrt{a_1} + \sqrt{a_{n-1}}}

Show Solution

Question 4

Slot-2

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

20

18

21

19

Show Solution

Question 5

Slot-2

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

Show Solution

Question 6

Slot-2

Let a1,a2,… be integers such that a1−a2+a3−a4+⋯+(−1)n−1an=n,for n≥1.\text{Let } a_1, a_2, \dots \text{ be integers such that } a_1 - a_2 + a_3 - a_4 + \dots + (-1)^{n-1} a_n = n, \quad \text{for } n \geq 1. Then a51+a52+⋯+a1023 equals ?\text{Then } a_{51} + a_{52} + \dots + a_{1023} \text{ equals ?}

-1

1

0

10

Show Solution

Question 7

Slot-1

A new species lays exactly 120 eggs out of which 50% are male and 50% are female. The female insect hatch and grow in a span of 20 days to lay eggs by themselves. On 1st April, 2018, an insect laid 120 eggs. Find how many eggs will be hatched, (approx) by the end of may 2018.

216000

25920000

432000

None of these

Show Solution

Question 8

Slot-1

Amar consumed 100 laddoos from Monday to Friday. Each day, he consumed 6 more laddoos than the previous day. How many laddoos did he consume on Wednesday?

12 laddoos

20 laddoos

24 laddoos

8 laddoos

Show Solution

Question 9

Slot-1

When opening his fruit shop for the day a shopkeeper found that his stock of apples could be perfectly arranged in a complete triangular array: that is, every row with one apple more than the row immediately above, going all the way up ending with a single apple at the top.

During any sales transaction, apples are always picked from the uppermost row, and going below only when that row is exhausted.

When one customer walked in the middle of the day she found an incomplete array in display having 126 apples totally. How many rows of apples (complete and incomplete) were seen by this customer? (Assume that the initial stock did not exceed 150 apples.)

15

14

13

12

11

Show Solution

CAT 2018 Progression questions

Question 1

Slot-1

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 an arithmetic progression then the common ratio of the geometric progression is

16\frac{1}{6}

36\frac{3}{6}

32\frac{3}{2}

52\frac{5}{2}

Show Solution

Question 2

Slot-2

The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ..... + 95 x 99 is

80707

80751

80730

80773

Show Solution

Question 3

Slot-1

Simplify the given expression: 1 + 1/2 + 1/(2²) + 1/(2³) + ... + 1/(2^(n-1)) = ?

(2^n - 1)/2^(n-1)

(2^n - 2)/2^(n-1)

2 - 2^n

2^n / (2^n-1)

Show Solution

Question 4

Slot-1

David has an interesting habit of spending money. He spends exactly £XX on the XXth day of a month. For example, he spends exactly £55 on the 55th of any month. On a few days in a year, David noticed that his cumulative spending during the last 44 consecutive days can be expressed as 2N2^N where NN is a natural number. What can be the possible value(s) of NN?

5

6

7

8

NN can have more than one value

Show Solution

Question 5

Slot-1

An antique store has a collection of 88 clocks. At a particular moment, the displayed times on 77 of the 88 clocks were as follows: 1:551:55 pm, 2:032:03 pm, 2:112:11 pm, 2:242:24 pm, 2:452:45 pm, 3:193:19 pm and 4:144:14 pm. If the displayed times of all 88 clocks form a mathematical series, then what was the displayed time on the remaining clock?

1:53 pm

1:58 pm

2:18 pm

3:08 pm

5:08 pm

Show Solution

CAT 2017 Progression questions

Question 1

Slot-1

Let a1,a2,…,a3na_1, a_2, \ldots, a_{3n} be an arithmetic progression with a1=3a_1 = 3 and a2=7a_2 = 7. If a1+a2+…+a3n=1830a_1 + a_2 + \ldots + a_{3n} = 1830, then what is the smallest positive integer mm such that m(a1+a2+…+an)>1830m(a_1 + a_2 + \ldots + a_n) > 1830?

8

9

10

11

Show Solution

Question 2

Slot-1

If the square of the 7th7^{\text{th}} term of an arithmetic progression with positive common difference equals the product of the 3rd3^{\text{rd}} and 17th17^{\text{th}} terms, then the ratio of the first term to the common difference is:

2 : 3

3 : 2

3 : 4

4 : 3

Show Solution

Question 3

Slot-2

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

132\frac{1}{32}

232\frac{2}{32}

332\frac{3}{32}

432\frac{4}{32}

Show Solution

Question 4

Slot-2

If a 1 = 12×5\frac{1}{2 × 5} , a 2 = 15×8\frac{1}{5 × 8} , a 3 = 18×11\frac{1}{8 × 11} ,...., then a 1 + a 2 + a 3 + ...... + a 100 is

25151\frac{25}{151}

12\frac{1}{2}

14\frac{1}{4}

11155\frac{111}{55}

Show Solution

Question 5

Slot-1

The sum of series, (−100)+(−95)+(−90)+⋯+110+115+120(-100) + (-95) + (-90) + \cdots + 110 + 115 + 120, is:

0

220

340

450

None of the above

Show Solution

Question 6

Slot-1

Find out next number in the series 2, 7, 17, 34, 60, ......

97

108

95

121

Show Solution
Practice all 646 Progression questionsFree, with worked solutions, beyond the previous year papers on this page.

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