Progressions, Sequences and Series Previous Year Questions (PYQs) for CAT: 5+ Solved Questions with Step-by-Step Solutions

Solve 5+ Progressions, Sequences and Series previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Progressions, Sequences and Series

Chapter Roadmap

Progressions, Sequences and Series

CAT tests whether you can convert a pattern into a formula, a sum, or a hidden equality.

➕

t1. Arithmetic Progressions

Master common difference, nth term, sums, common terms, and balance tricks.

4 PYQs
higher weight

✖️

t2. Geometric Progressions and Exponential Sequences

Handle constant ratio patterns, exponential terms, and compact sum logic.

2 PYQs
moderate

By the end: you should be able to look at a sequence and quickly decide whether to use term formula, sum formula, common-term logic, or symmetry.

Arithmetic Progressions

Selected Topic

Arithmetic Progressions

A constant step creates a predictable sequence.

Chapter: Progressions, Sequences and Series Topic t1 4 direct CAT PYQs

What you will learn here

Identify APs using common difference
Use nth term and sum formulas
Exploit symmetry of equally spaced terms
Find common terms of two APs
Solve CAT balance and hidden-equation patterns

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

Question 1 · Quantitative Ability MCQ

Suppose $x_{1}, x_{2}, x_{3}, ..., x_{100}$ are in arithmetic progression such that $x_{5} = - 4$ and $2 x_{6} + 2 x_{9} = x_{11} + x_{13}$ , Then, $x_{100}$ equals

A.
-194
B.
-196
C.
204
D.
206

Correct Answer:

Step-by-Step Solution

This is an AP term-equation question. You can recognise it because we are given one value of a term ( $x_{5} = - 4$ ) and a relation connecting several other terms ( $2 x_{6} + 2 x_{9} = x_{11} + x_{13}$ ) — the standard method is to write every term using $x_{n} = a + (n - 1) d$ and turn both pieces of information into two equations in $a$ and $d$ .

Step 1: Write the relation in terms of $a$ and $d$ .

$x_{6} = a + 5 d, x_{9} = a + 8 d, x_{11} = a + 10 d, x_{13} = a + 12 d$

So $2 x_{6} + 2 x_{9} = 2 (a + 5 d) + 2 (a + 8 d) = 4 a + 26 d$

and $x_{11} + x_{13} = (a + 10 d) + (a + 12 d) = 2 a + 22 d$

Step 2: Set them equal (as given) and simplify.

$4 a + 26 d = 2 a + 22 d$

$2 a + 4 d = 0$

$a + 2 d = 0 \Rightarrow a = - 2 d$

Step 3: Use the known value $x_{5} = - 4$ .

$x_{5} = a + 4 d = - 4$

Substitute $a = - 2 d$ :

$- 2 d + 4 d = - 4$

$2 d = - 4 \Rightarrow d = - 2$

Step 4: Find $a$ .

$a = - 2 d = - 2 (- 2) = 4$

Step 5: Find $x_{100}$ using the nth-term formula (careful: it's $n - 1$ , not $n$ ).

$x_{100} = a + 99 d = 4 + 99 (- 2) = 4 - 198 = - 194$

So $x_{100} = - 194$ .

Common trap: writing $x_{100} = a + 100 d$ instead of $a + 99 d$ gives $4 - 200 = - 196$ , which is exactly one of the wrong options — always double check you used $(n - 1)$ .

Question 2 · Quantitative Ability MCQ

Let $a_{n} = 46 + 8 n$ and $b_{n} = 98 + 4 n$ be two sequences for natural numbers $n \leq 100$ . Then, the sum of all terms common to both the sequences is

A.
14900
B.
14798
C.
15000
D.
14602

Correct Answer:

Step-by-Step Solution

This is a "common terms of two arithmetic progressions" question. The signal is two sequence formulas given in terms of $n$ , and you're asked for something about the numbers that appear in BOTH sequences.

Step 1: Write out the two sequences.

$a_{n} = 46 + 8 n$ for $n = 1, 2, \dots, 100$ : terms run from $54$ (at $n = 1$ ) to $846$ (at $n = 100$ ), increasing by 8 each time.

$b_{n} = 98 + 4 n$ for $n = 1, 2, \dots, 100$ : terms run from $102$ (at $n = 1$ ) to $498$ (at $n = 100$ ), increasing by 4 each time.

Step 2: Set the terms equal to find when a value appears in both.

A number $x$ is common if $x = 46 + 8 n = 98 + 4 m$ for some valid $n$ and $m$ .

Solve for $m$ in terms of $n$ :

$m = \frac{46 + 8 n - 98}{4} = \frac{8 n - 52}{4} = 2 n - 13$

Step 3: Apply the range restriction on $m$ .

Since $m$ must satisfy $1 \leq m \leq 100$ :

$1 \leq 2 n - 13 \leq 100$

$14 \leq 2 n \leq 113$

$7 \leq n \leq 56.5$

Since $n$ must be a whole number, $n$ ranges from $7$ to $56$ — that's $56 - 7 + 1 = 50$ values.

Step 4: Find the first and last common terms.

At $n = 7$ : $x = 46 + 8 (7) = 46 + 56 = 102$

At $n = 56$ : $x = 46 + 8 (56) = 46 + 448 = 494$

Step 5: Sum this arithmetic progression of common terms.

The common terms themselves form an AP (with common difference $8$ , since they're a subset of the $a_{n}$ sequence). Using the sum formula:

$Sum = \frac{number of terms}{2} \times (first + last) = \frac{50}{2} \times (102 + 494) = 25 \times 596 = 14900$

So the sum of all common terms is $14900$ .

Common trap: forgetting to apply BOTH range restrictions ( $1 \leq n \leq 100$ AND $1 \leq m \leq 100$ ) — using only one sequence's limit gives a wrong term count and a wrong sum.

Question 3 · Quantitative Ability MCQ

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

A.
41
B.
39
C.
43
D.
37

Correct Answer:

Step-by-Step Solution

This is a consecutive-odd-numbers balance-point question. Recognise it because we're told a value k inside a list of consecutive odd numbers splits the list into two equal-sum halves — this always uses the sum-of-odd-numbers shortcut and a bit of position algebra.

Step 1: List the sequence and count terms.

${1, 3, 5, \dots, 57}$ has $\frac{57 - 1}{2} + 1 = 29$ terms. The sum of the first $n$ odd numbers is $n^{2}$ , so the total sum is $2 9^{2} = 841$ .

Step 2: Represent k by its position.

Let $k$ be the $m$ -th odd number, so $k = 2 m - 1$ . The terms less than $k$ are the first $(m - 1)$ odd numbers, so their sum is $(m - 1)^{2}$ .

Step 3: Write the sum of terms greater than k.

Total $=$ (sum less than k) $+ k +$ (sum greater than k)

$841 = (m - 1)^{2} + (2 m - 1) + (sum greater)$

So sum greater $= 841 - (m - 1)^{2} - (2 m - 1)$ .

Step 4: Set sum-less equal to sum-greater (the given condition).

$(m - 1)^{2} = 841 - (m - 1)^{2} - (2 m - 1)$

$2 (m - 1)^{2} = 842 - 2 m$

$(m - 1)^{2} = 421 - m$

$m^{2} - 2 m + 1 = 421 - m$

$m^{2} - m - 420 = 0$

Step 5: Solve the quadratic.

$m = \frac{1 \pm 1 + 1680}{2} = \frac{1 \pm 41}{2}$

Taking the positive root, $m = 21$ .

Step 6: Find k.

$k = 2 m - 1 = 2 (21) - 1 = 41$

So $k = 41$ .

Common trap: a sign slip while expanding $(m - 1)^{2}$ (writing $m^{2} + 2 m + 1$ instead of $m^{2} - 2 m + 1$ ) flips the quadratic to $m^{2} + m - 420 = 0$ , giving $m = 20$ and the wrong $k = 39$ .

Question 4 · Quantitative Ability NAT

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

Correct Answer:

Step-by-Step Solution

Pattern: two regular polygons linked by a side-count ratio and an interior-angle ratio. Whenever you see "sides in ratio ... angles in ratio ...", set the number of sides as variables tied by the given ratio, write the interior angle formula for each, and solve one equation.

Why this method: the phrase "number of sides in the ratio 1 : 2" tells us to write the side counts as $n$ and $2 n$ (not two independent unknowns) — this collapses the problem to a single variable $n$ .

Step 1 — Set up the sides.

Let polygon A have $n$ sides. Since the ratio of sides is $1 : 2$ , polygon B has $2 n$ sides.

Step 2 — Write the interior angle formula.

For a regular polygon with $k$ sides, interior angle $= \frac{( k - 2 ) \times 18 0 ^{\circ}}{k}$ .

So:

$I_{A} = \frac{( n - 2 ) \times 180}{n}, I_{B} = \frac{( 2 n - 2 ) \times 180}{2 n} = \frac{( n - 1 ) \times 180}{n}$

Step 3 — Use the angle ratio.

$\frac{I _{A}}{I _{B}} = \frac{3}{4} ⟹ \frac{\frac{( n - 2 ) \times 180}{n}}{\frac{( n - 1 ) \times 180}{n}} = \frac{3}{4}$

The $180$ and $n$ cancel out top and bottom:

$\frac{n - 2}{n - 1} = \frac{3}{4}$

Step 4 — Solve for n.

$4 (n - 2) = 3 (n - 1) ⟹ 4 n - 8 = 3 n - 3 ⟹ n = 5$

Step 5 — Answer the actual question.

The question asks for the number of sides of B, which is $2 n = 2 \times 5 = 10$ .

Check: interior angle of a 5-gon $= \frac{3 \times 180}{5} = 10 8^{\circ}$ . Interior angle of a 10-gon $= \frac{8 \times 180}{10} = 14 4^{\circ}$ . Ratio $108 : 144 = 3 : 4$ ✓, and sides $5 : 10 = 1 : 2$ ✓.

Common trap: stopping at $n = 5$ and writing that as the final answer — the question wants the sides of B, which is $2 n$ , not $n$ .

Final answer: 10

Question 5 · Quantitative Ability MCQ

Let $a_{n}$ be the $n^{t h}$ term of a decreasing infinite geometric progression. If $a_{1} + a_{2} + a_{3} = 52$ and $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{1} = 624$ , then the sum of this geometric progression is

A.
57
B.
54
C.
60
D.
63

Correct Answer:

Step-by-Step Solution

Pattern: a regular octagon with a square formed by joining alternate vertices. Recognise it from "regular octagon" + naming every other vertex (A, C, E, G) — this is the classic "corner-cut square" construction.

Why this method: a regular octagon can always be seen as a square with its four corners cut off by 45°-45°-90° triangles. Alternate vertices of the octagon then sit exactly at the points where those cuts happen, so their square can be found with coordinates built from the corner-cut picture.

Step 1 — Build the octagon from a square.

Let each cut-off corner be an isosceles right triangle with legs of length $t$ . Its hypotenuse becomes one side of the octagon, so:

$a = t 2 ⟹ t = \frac{a}{2}$

If the original square has side $L$ , the leftover flat segment on each side of the square (which is also an octagon side) is $L - 2 t$ , and this must also equal $a$ :

$L - 2 t = a ⟹ L = a + 2 t = a + a 2 = a (1 + 2)$

Step 2 — Place coordinates.

Put the square with corners at $(0, 0), (L, 0), (L, L), (0, L)$ . After cutting the corners, label octagon vertices going around, including:

$A = (t, 0), C = (L, t)$

(A and C are exactly two vertices apart — alternate vertices — going around the octagon.)

Step 3 — Find AC, a side of square ACEG.

$A C^{2} = (L - t)^{2} + t^{2}$

Using $t = a / 2$ and $L - t = a (1 + 2) - a / 2 = a (1 + \frac{2}{2})$ :

$A C^{2} = a^{2} (1 + \frac{2}{2})^{2} + a^{2} (\frac{1}{2})^{2}$

$= a^{2} (1 + 2 + \frac{1}{2}) + a^{2} \cdot \frac{1}{2} = a^{2} (2 + 2)$

Step 4 — Plug in a = 6.

$Area of ACEG = A C^{2} = 6^{2} (2 + 2) = 36 (2 + 2)$

Common trap: confusing this alternate-vertex square with the bounding square of side $L = a (1 + 2)$ — that gives $L^{2} = a^{2} (1 + 2)^{2} = a^{2} (3 + 22)$ , a completely different (and larger) quantity, and is not one of the listed shortcuts either; students often instead misremember the coefficient and pick $a^{2} (1 + 2)$ by mistake.

Final answer: 36(2+√2), option D

Progressions, Sequences and Series Previous Year Questions (PYQs) for CAT: 5+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Progressions, Sequences and Series

Progressions, Sequences and Series

Arithmetic Progressions

Arithmetic Progressions

Progressions, Sequences and Series: 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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