Progressions, Sequences and Series Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Progressions, Sequences and Series notes for CAT: 36 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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

AP Intuition: Same Jump Every Time

An AP is a sequence with equal jumps

If every term is obtained by adding the same number to the previous term, the sequence is an arithmetic progression.

The repeated jump is called the common difference, usually denoted by

d

The nth Term Formula

Term position = number of jumps

The first term has made zero jumps. The nth term has made $n - 1$ jumps.

a_{n} = a + (n - 1) d

$a$ : first term

$d$ : common difference

$n - 1$ : number of jumps from the first term

Progressions, Sequences and Series: Solved Questions with Step-by-Step Explanations (2 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.

Progressions, Sequences and Series Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Chapter Roadmap: Progressions, Sequences and Series

Progressions, Sequences and Series

Arithmetic Progressions

Arithmetic Progressions

AP Intuition: Same Jump Every Time

An AP is a sequence with equal jumps

The nth Term Formula

Term position = number of jumps

Progressions, Sequences and Series: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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