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

Progressions, Sequences and Series short notes for CAT: 33 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

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

Sum of First n Terms

Sum = number of terms × average term

In an AP, the average of all terms equals the average of the first and last term.

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

S_{n} = \frac{n}{2} {2 a + (n - 1) d}

Use the first formula when the last term is known. Use the second formula when

a

and

d

are known.

Equally Spaced Terms Have a Middle Average

Symmetry is a shortcut

Terms equally spaced around a middle term have the middle term as their average.

4th term

a + 3 d

↔

7th term

a + 6 d

↔

10th term

a + 9 d

a_{4} + a_{7} + a_{10} = 3 a_{7}

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.

More short notes in this unit

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

Progressions, Sequences and Series short notes for CAT: 33 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

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 ?

Question 2

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?

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