Sequences, Series and Progressions Previous Year Questions (PYQs) for CAT: 14+ Solved Questions with Step-by-Step Solutions

Solve 14+ Sequences, Series and Progressions previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Sequences, Series and Progressions

Chapter roadmap

Sequences, Series and Progressions

➕ Arithmetic Progressions and Common Terms

Master fixed-difference sequences, AP sums, average of terms, common terms, and AP-based integer conditions.

6 direct CAT PYQs | selected and strongest topic

🔁 Recursive Sequences and Patterned Terms

Learn how terms depend on earlier terms and how to detect hidden cycles or telescoping behavior.

4 direct CAT PYQs

Σ Series Summation and Infinite Series

Convert long sums into compact forms using structure, grouping, and infinite-series logic.

2 direct CAT PYQs

📈 Growth Sequences and Applied Recurrences

Apply sequence logic to growth, grouping, experiments, and word-problem recurrence models.

4 direct CAT PYQs

By the end, you should see whether a question is asking for a term, a sum, a common term, or a hidden pattern.

Topic Hero: Arithmetic Progressions and Common Terms

Algebra → Sequences, Series and Progressions → Topic 1

Same jump, every time

Arithmetic Progressions and Common Terms

CAT often hides clean linear patterns inside terms, sums, averages, and common-term conditions.

✅ Find the

n

th term of an AP

✅ Use AP sum and average shortcuts

✅ Decode AP from sum of first

n

terms

✅ Solve common terms of two APs

✅ Handle three integers in AP

Sequences, Series and Progressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

Let both the series $a_{1}, a_{2}, a_{3} ...$ and $b_{1}, b_{2}, b_{3} ...$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_{5} = b_{9}$ , $a_{19} = b_{19}$ and $b_{2} = 0$ , then $a_{11}$ equals

A.
86
B.
79
C.
83
D.
84

Correct Answer:

Step-by-Step Solution

Key idea: this is a two-AP matching-terms question. We are told that two arithmetic progressions have prime common differences and that certain terms are equal. The fastest method is to write the general term of each AP and convert the given equalities into equations.

Step 1: Let

$a_{n} = A + (n - 1) p$

and

$b_{n} = B + (n - 1) q,$

where $p$ and $q$ are the common differences. Both $p$ and $q$ are prime numbers.

Step 2: Use $b_{2} = 0$ :

$b_{2} = B + q = 0$

$B = - q .$

Step 3: Use $a_{5} = b_{9}$ :

$A + 4 p = B + 8 q .$

Substitute $B = - q$ :

$A + 4 p = - q + 8 q = 7 q .$

Hence

$A + 4 p = 7 q . (1)$

Step 4: Use $a_{19} = b_{19}$ :

$A + 18 p = B + 18 q .$

Substitute $B = - q$ :

$A + 18 p = - q + 18 q = 17 q .$

Hence

$A + 18 p = 17 q . (2)$

Step 5: Subtract equation (1) from equation (2):

$(A + 18 p) - (A + 4 p) = 17 q - 7 q$

$14 p = 10 q .$

Divide by 2:

$7 p = 5 q .$

Step 6: Since $p$ and $q$ are primes, the only positive prime solution is

$p = 5, q = 7.$

Step 7: Find $A$ using equation (1):

$A + 4 (5) = 7 (7)$

$A + 20 = 49$

$A = 29.$

Step 8: Now compute $a_{11}$ :

$a_{11} = A + 10 p = 29 + 10 (5) = 29 + 50 = 79.$

Answer: B

Question 2 · Quantitative Ability MCQ

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

A.
6119
B.
6090
C.
4941
D.
7471

Correct Answer:

Step-by-Step Solution

Key idea: This is a grouped sequence question where natural numbers are partitioned into groups of increasing odd sizes.

Step 1: Identify the pattern in group sizes and last terms.

Group 1: (1) → 1 term, ends at $1 = 1^{2}$

Group 2: (2, 3, 4) → 3 terms, ends at $4 = 2^{2}$

Group 3: (5, 6, 7, 8, 9) → 5 terms, ends at $9 = 3^{2}$

The $n$ -th group has $2 n - 1$ terms and its last term is $n^{2}$ .

Step 2: Find the first term of the $n$ -th group.

Since the previous group ended at $(n - 1)^{2}$ , the current group starts at $(n - 1)^{2} + 1$ .

Step 3: Calculate the sum of the $n$ -th group.

The terms are consecutive integers, forming an Arithmetic Progression (AP) with common difference $d = 1$ .

Sum = $\frac{number of terms}{2} \times (first term + last term)$

$S_{n} = \frac{2 n - 1}{2} \times ((n - 1)^{2} + 1 + n^{2})$

Step 4: Substitute $n = 15$ to find the sum of the 15th group.

Number of terms = $2 (15) - 1 = 29$ .

First term = $(15 - 1)^{2} + 1 = 1 4^{2} + 1 = 196 + 1 = 197$ .

Last term = $1 5^{2} = 225$ .

Sum = $\frac{29}{2} \times (197 + 225) = \frac{29}{2} \times 422 = 29 \times 211$ .

Step 5: Compute the final product.

$29 \times 211 = 29 \times (200 + 10 + 1) = 5800 + 290 + 29 = 6119$ .

Answer: 6119

Question 3 · Quantitative Ability MCQ

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

A.
9
B.
4
C.
7
D.
8

Correct Answer:

Step-by-Step Solution

Key idea: This is an AP sum-to-term question. It is recognisable because the sum of the first $n$ terms is given, and the question asks about the $n$ th term.

Step 1: Write the given sum formula.

$S_{n} = n + 2 n^{2} = 2 n^{2} + n$

Step 2: Use the relation between sum and term.

For any progression,

$T_{n} = S_{n} - S_{n - 1}$

This is the safest method because it does not require finding the first term and common difference separately.

Step 3: Compute $S_{n - 1}$ .

$S_{n - 1} = 2 (n - 1)^{2} + (n - 1)$

Expand carefully:

$2 (n - 1)^{2} = 2 (n^{2} - 2 n + 1) = 2 n^{2} - 4 n + 2$

So,

$S_{n - 1} = 2 n^{2} - 4 n + 2 + n - 1 = 2 n^{2} - 3 n + 1$

Step 4: Subtract to get the $n$ th term.

$T_{n} = (2 n^{2} + n) - (2 n^{2} - 3 n + 1)$

$T_{n} = 2 n^{2} + n - 2 n^{2} + 3 n - 1 = 4 n - 1$

Step 5: Apply the divisibility condition.

We need $T_{n}$ to be divisible by 9:

$4 n - 1 \equiv 0 (mod 9)$

$4 n \equiv 1 (mod 9)$

The inverse of 4 modulo 9 is 7, because $4 \times 7 = 28 \equiv 1 (mod 9)$ .

Therefore,

$n \equiv 7 (mod 9)$

Step 6: Choose the smallest natural number.

The smallest positive $n$ satisfying this is $n = 7$ .

Answer: 7

Question 4 · Quantitative Ability NAT

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

Correct Answer:

967

Step-by-Step Solution

Key idea: this is a common-terms question between two APs. Each AP can be written as a congruence condition, and common terms occur where both congruences are satisfied.

Step 1: Write the first sequence:

$a_{n} = 13 + 6 (n - 1) .$

Its terms are

$13, 19, 25, 31, \dots$

Every term satisfies

$x \equiv 13 (mod 6) .$

Since $13 \equiv 1 (mod 6)$ , this becomes

$x \equiv 1 (mod 6) .$

Step 2: Write the second sequence:

$b_{n} = 15 + 7 (n - 1) .$

Its terms are

$15, 22, 29, 36, \dots$

Every term satisfies

$x \equiv 15 (mod 7) .$

Since $15 \equiv 1 (mod 7)$ , this becomes

$x \equiv 1 (mod 7) .$

Step 3: A common term must satisfy both:

$x \equiv 1 (mod 6)$

and

$x \equiv 1 (mod 7) .$

Since 6 and 7 are coprime, this means

$x \equiv 1 (mod 42) .$

Step 4: The numbers congruent to 1 modulo 42 are

$1, 43, 85, 127, \dots$

The first one that actually appears in both given sequences is 43. Therefore the common terms form the AP

$43, 85, 127, \dots$

with common difference 42.

Step 5: We need the largest three-digit common term. Let it be

$43 + 42 t \leq 999.$

Then

$42 t \leq 956.$

The largest integer $t$ is 22.

Step 6: Compute the term:

$43 + 42 (22) = 43 + 924 = 967.$

Answer: 967

Question 5 · Quantitative Ability MCQ

For a sequence of real numbers $x_{1}, x_{2}, ... x_{n}$ , If $x_{1} - x_{2} + x_{3} - .... + (- 1)^{n + 1} x_{n} = n^{2} + 2 n$ for all natural numbers n, then the sum $x_{49} + x_{50}$ equals

A.
200
B.
2
C.
-200
D.
-2

Correct Answer:

Step-by-Step Solution

Key idea: this is a running-alternating-sum question. The expression given for all $n$ is the alternating sum up to $x_{n}$ . To recover individual terms, subtract consecutive partial sums.

Step 1: Define

$S_{n} = x_{1} - x_{2} + x_{3} - \dots + (- 1)^{n + 1} x_{n} .$

The question gives

$S_{n} = n^{2} + 2 n .$

Step 2: For $n \geq 2$ ,

$S_{n} - S_{n - 1} = (- 1)^{n + 1} x_{n} .$

This is because all terms up to $x_{n - 1}$ cancel, leaving only the signed nth term.

Step 3: Compute $S_{n - 1}$ :

$S_{n - 1} = (n - 1)^{2} + 2 (n - 1) .$

Expand:

$S_{n - 1} = n^{2} - 2 n + 1 + 2 n - 2 = n^{2} - 1.$

Step 4: Subtract:

$S_{n} - S_{n - 1} = (n^{2} + 2 n) - (n^{2} - 1) .$

$S_{n} - S_{n - 1} = 2 n + 1.$

Therefore,

$(- 1)^{n + 1} x_{n} = 2 n + 1.$

Step 5: Solve for $x_{n}$ :

$x_{n} = (- 1)^{n + 1} (2 n + 1) .$

Step 6: Find $x_{49}$ . Since 49 is odd,

$(- 1)^{50} = 1,$

$x_{49} = 2 (49) + 1 = 99.$

Step 7: Find $x_{50}$ . Since 50 is even,

$(- 1)^{51} = - 1,$

$x_{50} = - (2 (50) + 1) = - 101.$

Step 8: Add:

$x_{49} + x_{50} = 99 + (- 101) = - 2.$

Answer: D

Sequences, Series and Progressions Previous Year Questions (PYQs) for CAT: 14+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Sequences, Series and Progressions

Sequences, Series and Progressions

Topic Hero: Arithmetic Progressions and Common Terms

Arithmetic Progressions and Common Terms

Sequences, Series and Progressions: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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