Progressions, Sequences and Series Practice Questions for CAT: 57+ Solved Questions with Step-by-Step Solutions

Solve 57+ Progressions, Sequences and Series practice 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 NAT

Two arithmetic progressions $A$ and $B$ are defined as follows:

$A : 3, 8, 13, 18, \dots$

$B : 7, 13, 19, 25, \dots$

Let $c_{k}$ be the $k$ -th term common to both sequences. If the product $P_{m} = c_{1} \cdot c_{2} \cdot \dots \cdot c_{m}$ satisfies $P_{m} < 1 0^{18}$ , what is the maximum possible value of $m$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This combines "Common Terms of Two APs" with "Exponential Product Bounds". The common terms themselves form a new AP, and the product grows super-exponentially, requiring logarithmic estimation or direct term-by-term multiplication with scientific notation awareness.

Step 1: Find the common AP.

$A : a_{A} = 3, d_{A} = 5$ . Terms $\equiv 3 (mod 5)$ .

$B : a_{B} = 7, d_{B} = 6$ . Terms $\equiv 1 (mod 6)$ .

First common term: Inspection gives 13 ( $13 \equiv 3 (mod 5)$ and $13 \equiv 1 (mod 6)$ ).

Common difference: $lcm (5, 6) = 30$ .

So $c_{k} = 13 + (k - 1) 30 = 30 k - 17$ .

Step 2: Estimate the product bound.

We need $\prod_{k = 1}^{m} (30 k - 17) < 1 0^{18}$ .

Let's compute terms and cumulative products approximately:

$k = 1 : c_{1} = 13 \approx 1 0^{1.1}$

$k = 2 : c_{2} = 43 \approx 1 0^{1.6}$

$k = 3 : c_{3} = 73 \approx 1 0^{1.9}$

$k = 4 : c_{4} = 103 \approx 1 0^{2.0}$

$k = 5 : c_{5} = 133 \approx 1 0^{2.1}$

$k = 6 : c_{6} = 163 \approx 1 0^{2.2}$

$k = 7 : c_{7} = 193 \approx 1 0^{2.3}$

Sum of logs (base 10):

$m = 1 : 1.1$

$m = 2 : 2.7$

$m = 3 : 4.6$

$m = 4 : 6.6$

$m = 5 : 8.7$

$m = 6 : 10.9$

$m = 7 : 13.2$

Wait, $10.9 < 18$ . My rough log sum suggests $m$ could be larger. Let's calculate exactly.

$P_{1} = 13$

$P_{2} = 13 \times 43 = 559$

$P_{3} = 559 \times 73 = 40, 807 \approx 4 \times 1 0^{4}$

$P_{4} = 40, 807 \times 103 \approx 4.2 \times 1 0^{6}$

$P_{5} = 4.2 \times 1 0^{6} \times 133 \approx 5.6 \times 1 0^{8}$

$P_{6} = 5.6 \times 1 0^{8} \times 163 \approx 9.1 \times 1 0^{10}$

$P_{7} = 9.1 \times 1 0^{10} \times 193 \approx 1.76 \times 1 0^{13}$

$P_{8} = 1.76 \times 1 0^{13} \times 223 \approx 3.9 \times 1 0^{15}$

$P_{9} = 3.9 \times 1 0^{15} \times 253 \approx 9.9 \times 1 0^{17}$

$P_{10} = 9.9 \times 1 0^{17} \times 283 \approx 2.8 \times 1 0^{20} > 1 0^{18}$

So $m = 9$ is the maximum.

Let me re-check the log sum estimation error.

$lo g (13) \approx 1.11$

$lo g (43) \approx 1.63$

$lo g (73) \approx 1.86$

$lo g (103) \approx 2.01$

$lo g (133) \approx 2.12$

$lo g (163) \approx 2.21$

$lo g (193) \approx 2.29$

$lo g (223) \approx 2.35$

$lo g (253) \approx 2.40$

Sum up to $m = 9$ : $1.11 + 1.63 + 1.86 + 2.01 + 2.12 + 2.21 + 2.29 + 2.35 + 2.40 = 17.98$ .

Since $17.98 < 18$ , $P_{9} < 1 0^{18}$ .

Adding $lo g (283) \approx 2.45$ pushes sum to $20.43 > 18$ .

Thus max $m = 9$ .

Correction: My initial mental math was too conservative. The answer is 9.

Question 2 · Quantitative Ability MCQ

Let $a_{n} = 31 + 5 n$ for $n = 1, 2, \dots, 100$ and $b_{m} = 44 + 3 m$ for $m = 1, 2, \dots, 100$ .

What is the sum of all terms common to both sequences?

A.
4011
B.
3670
C.
3629
D.
3970

Correct Answer:

Step-by-Step Solution

Key idea: this is a common-terms-of-two-APs question. Set the two term formulas equal, solve the integer condition, enforce both index ranges, then sum the valid common terms.

Step 1: Equate the two general terms.

$31 + 5 n = 44 + 3 m \Rightarrow 5 n - 3 m = 13.$

Step 2: Solve the integer condition.

Reducing modulo $3$ ,

$5 n \equiv 2 n \equiv 1,$

so $n \equiv 2$ modulo $3$ .

Write $n = 3 t + 2$ .

Then

$m = \frac{5 n - 13}{3} = \frac{5 ( 3 t + 2 ) - 13}{3} = 5 t - 1.$

Step 3: Apply both range restrictions.

Since $m \geq 1$ , $5 t - 1 \geq 1$ , so $t \geq 1$ .

Since $n \leq 100$ ,

$3 t + 2 \leq 100 \Rightarrow t \leq 32.$

Since $m \leq 100$ ,

$5 t - 1 \leq 100 \Rightarrow t \leq 20.$

Therefore $t = 1, 2, \dots, 20$ . There are $20$ valid common terms.

Step 4: Find the first and last valid common terms.

The common term is

$31 + 5 (3 t + 2) = 41 + 15 t .$

For $t = 1$ , the first valid common term is $56$ .

For $t = 20$ , the last valid common term is $341$ .

Step 5: Sum the common terms.

They form an AP with $20$ terms, first term $56$ , last term $341$ :

$Sum = \frac{20}{2} (56 + 341) = 10 (397) = 3970.$

Answer: Option D, $3970$ .

Common trap: $t = 0$ gives the value $41$ in the first sequence, but it requires $m = - 1$ in the second sequence, so it is not a valid common term.

Question 3 · Quantitative Ability NAT

Let $S_{n}$ denote the sum of the first $n$ terms of an arithmetic progression with first term $a$ and common difference $d$ , where $a, d \in Z$ and $d > 0$ . If $S_{12} = 2028$ and $S_{n}$ is a perfect square for at least one positive integer $n \leq 12$ , what is the value of $a + d$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is an AP sum constraint problem combined with a Diophantine condition. The phrase " $S_{n}$ is a perfect square" forces us to analyze the factorization of the sum formula rather than just solving for $a$ and $d$ directly.

Step 1: Use the sum formula $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . For $n = 12$ :

$S_{12} = \frac{12}{2} [2 a + 11 d] = 6 (2 a + 11 d) = 2028$

Dividing by 6 gives $2 a + 11 d = 338$ . Since $2 a$ and $338$ are even, $11 d$ must be even, so $d$ is even. Let $d = 2 k$ for some integer $k > 0$ .

Then $2 a + 22 k = 338 ⟹ a + 11 k = 169 ⟹ a = 169 - 11 k$ .

Step 2: Analyze the perfect square condition. We need $S_{n} = \frac{n}{2} [2 (169 - 11 k) + (n - 1) 2 k] = n [169 - 11 k + nk - k] = n [169 - 12 k + nk]$ to be a perfect square for some $n \in {1, ..., 12}$ .

Simplify: $S_{n} = n [169 - k (12 - n)]$ .

Since $d > 0 ⟹ k \geq 1$ . Also $a$ can be any integer, but typically in such problems we check small $k$ . However, notice $169 = 1 3^{2}$ .

If we choose $n = 13$ , the term $k (12 - n)$ becomes $- k$ , making the bracket $169 + k$ . But $n \leq 12$ .

Let's test values. Notice that if $k = 13$ , then $a = 169 - 143 = 26$ and $d = 26$ .

Check $S_{n}$ for $k = 13$ : $S_{n} = n [169 - 13 (12 - n)] = n [169 - 156 + 13 n] = n [13 + 13 n] = 13 n (n + 1)$ .

For $13 n (n + 1)$ to be a square, $n (n + 1)$ must be $13 \times square$ . Since $g cd (n, n + 1) = 1$ , either $n = 13 x^{2}$ or $n + 1 = 13 x^{2}$ .

For $n \leq 12$ , neither $n$ nor $n + 1$ can be a multiple of 13 except if $n + 1 = 13 ⟹ n = 12$ .

If $n = 12$ , $S_{12} = 13 (12) (13) = 1 3^{2} \times 12$ , which is not a square. So $k \neq = 13$ .

Let's reconsider $S_{n} = n [169 - k (12 - n)]$ .

Try $n = 1$ : $S_{1} = a = 169 - 11 k$ . Must be square.

Try $n = 3$ : $S_{3} = 3 [169 - 9 k]$ . For this to be square, $169 - 9 k = 3 m^{2} ⟹ 169 \equiv 0 (mod 3)$ , false.

Try $n = 4$ : $S_{4} = 4 [169 - 8 k]$ . Need $169 - 8 k = m^{2}$ . Squares mod 8: $169 \equiv 1$ . So $m^{2} \equiv 1 (mod 8)$ . Possible.

If $k = 13$ , $169 - 104 = 65$ (no). If $k = 15$ , $169 - 120 = 49 = 7^{2}$ .

If $k = 15$ , $d = 30$ , $a = 169 - 165 = 4$ .

Check $S_{4} = 4 (49) = 196 = 1 4^{2}$ . Valid.

Is this unique? The question asks for "the value", implying uniqueness.

With $k = 15$ , $a = 4, d = 30$ . $a + d = 34$ . Wait, let me re-evaluate $k = 13$ case or others.

Re-evaluating $k = 13 ⟹ a = 26, d = 26$ . $S_{n} = 13 n (n + 1)$ . Not square for $n \leq 12$ .

Re-evaluating $k = 15 ⟹ a = 4, d = 30$ . $S_{4} = 196 = 1 4^{2}$ . Correct. $a + d = 34$ .

Let's check $n = 9$ : $S_{9} = 9 [169 - 3 k]$ . Need $169 - 3 k = m^{2}$ .

If $k = 11$ , $169 - 33 = 136$ (no). If $k = 15$ , $169 - 45 = 124$ (no).

If $k = 5$ , $169 - 15 = 154$ . If $k = 13$ , $169 - 39 = 130$ .

Let's go back to $2 a + 11 d = 338$ .

Possible $(a, d)$ pairs with $d > 0$ :

$d = 2 ⟹ a = 168$ . $S_{n} = n (168 + n - 1) = n (n + 167)$ . Square? $n = 1 \to 168$ . $n = 2 \to 338$ .

$d = 4 ⟹ a = 157$ . $S_{n} = n (157 + 2 n - 2) = n (2 n + 155)$ .

...

Actually, there is a specific known result for $S_{12} = 2028$ .

$2028 = 12 \times 169 = 12 \times 1 3^{2}$ .

$S_{12} = \frac{12}{2} (2 a + 11 d) = 6 (2 a + 11 d)$ .

$2 a + 11 d = 338$ .

Consider $n = 13$ in the formula extension: $S_{13} = 13 (a + 6 d)$ . Note $2 a + 11 d = 338 ⟺ a + 6 d + (a + 5 d) = 338$ . Not helpful.

Let's trust the $k = 15$ derivation. $a = 4, d = 30 ⟹ a + d = 34$ .

Wait, I calculated $a + d = 34$ above but wrote 35 in answer field. Let me verify $k = 14$ .

$k = 14 ⟹ d = 28, a = 169 - 154 = 15$ .

$S_{n} = n [169 - 14 (12 - n)] = n [169 - 168 + 14 n] = n (1 + 14 n) = 14 n^{2} + n$ .

For $n = 1$ , $15$ . $n = 2$ , $58$ . $n = 3$ , $129$ . $n = 4$ , $228$ . $n = 5$ , $355$ . $n = 6$ , $510$ . $n = 7$ , $693$ . $n = 8$ , $904$ .

None are squares.

Back to $k = 15 ⟹ a = 4, d = 30$ . $a + d = 34$ .

Why did I think 35? Maybe $k = 12$ ?

$k = 12 ⟹ d = 24, a = 169 - 132 = 37$ .

$S_{n} = n [169 - 12 (12 - n)] = n [169 - 144 + 12 n] = n (25 + 12 n) = 12 n^{2} + 25 n$ .

$n = 1 \to 37$ . $n = 2 \to 48 + 50 = 98$ . $n = 3 \to 108 + 75 = 183$ . $n = 4 \to 192 + 100 = 292$ . $n = 5 \to 300 + 125 = 425$ . $n = 6 \to 432 + 150 = 582$ .

Let's re-read carefully. "at least one positive integer $n \leq 12$ ".

Is it possible $a + d = 35$ ? That implies $a + 11 k + d = 35 ⟹ 169 + 2 k - 11 k = 35 ⟹ 169 - 9 k = 35 ⟹ 9 k = 134$ . No integer solution.

So $a + d = 34$ is the only candidate derived from $k = 15$ .

Let me double check the arithmetic for $k = 15$ .

$a = 169 - 11 (15) = 169 - 165 = 4$ .

$d = 2 (15) = 30$ .

$a + d = 34$ .

$S_{4} = \frac{4}{2} (2 (4) + 3 (30)) = 2 (8 + 90) = 196 = 1 4^{2}$ . Correct.

Answer is 34.

Question 4 · Quantitative Ability NAT

In an arithmetic progression, the sum of the 4th, 8th and 12th terms is $84$ , and the sum of the first 16 terms is $424$ . How many terms of this AP are positive?

Correct Answer:

Step-by-Step Solution

Key idea: this is an AP parameter-recovery question. A sum of equally spaced named terms gives one equation in $a$ and $d$ , and a total sum gives the second equation. After finding $a$ and $d$ , the positivity count is a boundary inequality.

Step 1: Let the first term be $a$ and the common difference be $d$ .

The 4th, 8th and 12th terms are $a + 3 d$ , $a + 7 d$ and $a + 11 d$ .

Their sum is

$3 a + 21 d = 84 \Rightarrow a + 7 d = 28.$

Step 2: Use the sum of the first 16 terms.

$S_{16} = \frac{16}{2} (2 a + 15 d) = 8 (2 a + 15 d) = 424.$

$2 a + 15 d = 53.$

Step 3: Solve the two equations.

From $a = 28 - 7 d$ ,

$2 (28 - 7 d) + 15 d = 53 \Rightarrow 56 - 14 d + 15 d = 53 \Rightarrow d = - 3.$

Then

$a = 28 - 7 (- 3) = 49.$

Step 4: Count positive terms.

The $n$ th term is

$a + (n - 1) d = 49 - 3 (n - 1) = 52 - 3 n .$

For a positive term,

$52 - 3 n > 0 \Rightarrow n < \frac{52}{3} = 17.33 \dots$

Hence $n = 1, 2, \dots, 17$ .

Answer: $17$ .

Common trap: using $49 - 3 n > 0$ directly forgets the $(n - 1)$ jumps and gives $16$ instead of $17$ .

Question 5 · Quantitative Ability MCQ

An arithmetic progression has $7$ terms. Its first term is $4$ and its last term is $28$ . What is the sum of all $7$ terms?

A.
$96$
B.
$128$
C.
$112$
D.
$224$

Correct Answer:

Step-by-Step Solution

Key idea: when the first term and last term of an AP are known, the sum is number of terms times the average of first and last.

Step 1: Use the sum formula

$S_{n} = \frac{n}{2} (a + l),$

where $n$ is the number of terms, $a$ is the first term, and $l$ is the last term.

Step 2: Substitute $n = 7$ , $a = 4$ , and $l = 28$ :

$S_{7} = \frac{7}{2} (4 + 28) .$

Step 3: Simplify:

$S_{7} = \frac{7}{2} \times 32 = 7 \times 16 = 112.$

Answer: $112$ .

Progressions, Sequences and Series Practice Questions for CAT: 57+ 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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