The number of common terms in the progressions 4, 9, 14, 19, up to 25th term and 3, 6, 9, 12, up to 37th term is :

Year	2026	2025	2024	Total
Questions	17	24	14	55

Q jee_main_2025_28_jan_morning Recurrence Relations and Summation

Let

< aₙ >

be a sequence such that

a₀ = 0, a₁ = (1)/(2)

and

2an + 2 = 5an + 1 - 3aₙ, n = 0, 1, 2, 3,

. Then

Σk = 1¹⁰⁰ ak

is equal to:

A. $3a₉₉ - 100$ $3a_{99} - 100$
B. $3a₁₀₀ - 100$ $3a_{100} - 100$
C. $3a₁₀₀ + 100$ $3a_{100} + 100$
D. $3a₉₉ + 100$ $3a_{99} + 100$

Solution

Related Formula

Characteristic equation method for standard second-order linear homogeneous recurrence updates:

2x² - 5x + 3 = 0

Core Logic

Solving the characteristic equation gives roots $x = 1$ $x = 1$ and $x = (3)/(2)$ $x = \frac{3}{2}$. The general solution takes the form:

aₙ = A(1)ⁿ + B((3)/(2))ⁿ

Step 1: Evaluating Sequence Parameters

Using boundary conditions: For $n = 0 A + B = 0$ $n = 0 \implies A + B = 0$ For $n = 1 A + (3)/(2)B = (1)/(2)$

Solving this simple linear system gives $B = 1$ $B = 1$ and $A = -1$ $A = -1$. Thus, the explicit sequence formula is:

aₙ = -1 + ((3)/(2))ⁿ

Step 2: Summing the Target Range

Σk = 1¹⁰⁰ ak = Σk = 1¹⁰⁰ (-1) + Σk = 1¹⁰⁰ ((3)/(2))^k

= -100 + (3)/(2)[((3)/(2))¹⁰⁰ - 1](3)/(2) - 1 = -100 + 3[((3)/(2))¹⁰⁰ - 1]

= 3a₁₀₀ - 100

Pattern Recognition

Characteristic roots directly decouple second-order linear loop progressions into basic combinations of clean geometric progressions.

Chapter Mix

Class 11 Maths: Sequences and Series

Q jee_main_2025_28_jan_morning Arithmetic Progression Properties

Let

Tᵣ

be the

rth

term of an A.P. If for some m,

Tm = (1)/(25), T₂₅ = (1)/(20)

and

20Σr = 1²⁵ Tᵣ = 13

then

5mΣr = m2m Tᵣ

is equal to:

A. 112
B. 126
C. 98
D. 142

Solution

Related Formula

Standard Arithmetic Progression summation template:

Sₙ = (n)/(2)[2a + (n-1)d]

Core Logic

Given structural constraints:

T₂₅ = a + 24d = (1)/(20)

20 \cdot (25)/(2)[a + (1)/(20)] = 13 a = (1)/(500)

Step 1: Finding Parameters and Indices

Substituting $a = (1)/(500)$ $a = \frac{1}{500}$ back into $a + 24d = (1)/(20)$ $a + 24d = \frac{1}{20}$ gives $d = (1)/(500)$ $d = \frac{1}{500}$.

Using the formula for $Tm$ $T_m$:

Tm = a + (m-1)d = (1)/(500) + (m-1)/(500) = (1)/(25) m = 20

Step 2: Computing the Target Segment Sum

For $m = 20$ $m = 20$, the target expression becomes:

5(20) Σr=20⁴⁰ Tᵣ = 100 \cdot (21)/(2) [T₂₀ + T₄₀]

Evaluating the values gives exactly 126.

Pattern Recognition

When $a = d$ $a = d$, the expressions simplify directly to basic multiples of the index position ( $Tₙ = n \cdot d$ $T_n = n \cdot d$), cutting down calculation time.

Chapter Mix

Class 11 Maths: Sequences and Series

Q jee_main_2025_03_april_morning Method of Differences

The sum

1 + 3 + 11 + 25 + 45 + 71 +

up to 20 terms, is equal to:

A. 7240
B. 7130
C. 6982
D. 8124

Solution

Related Formula

For a series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic expression:

Tₙ = an² + bn + c

Core Logic

Analyze the successive first-order differences of the terms:

Series: 1, 3, 11, 25, 45, 71

Differences: 2, 8, 14, 20, 26

Since the consecutive differences have a constant difference of $6$ $6$, they form an A.P. Set up the linear system for the first three terms:

$T₁ = a + b + c = 1$ $T_1 = a + b + c = 1$
$T₂ = 4a + 2b + c = 3$ $T_2 = 4a + 2b + c = 3$
$T₃ = 9a + 3b + c = 11$ $T_3 = 9a + 3b + c = 11$

Solving these equations simultaneously yields:

a = 3, b = -7, c = 5

Step 1: Summing the Series

The general term is:

Tₙ = 3n² - 7n + 5

Evaluate the summation for $n = 20$ $n = 20$ terms:

S₂₀ = Σn=1²⁰ (3n² - 7n + 5) = 3Σn=1²⁰ n² - 7Σn=1²⁰ n + Σn=1²⁰ 5

Substitute standard power sum formulas:

S₂₀ = 3 \cdot ((20 \cdot 21 \cdot 41)/(6)) - 7 \cdot ((20 \cdot 21)/(2)) + 5(20)

S₂₀ = 8610 - 1470 + 100 = 7240

Pattern Recognition

Shortcut: When the first-order differences form an arithmetic sequence, the $n$ $n$-th term is quadratic ( $an² + bn + c$ $an^2 + bn + c$) where the second difference is $2a$ $2a$ ( $2a = 6 a = 3$ $2a = 6 \implies a = 3$). Determine $b$ $b$ and $c$ $c$ using small values of $n$ $n$, then apply standard $Σ n²$ $\sum n^2$ and $Σ n$ $\sum n$ summation formulas.

Evaluation Rubric / Model Answer

7240

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q66 jee_main_2025_03_april_morning Geometric Progression

Let

a₁, a₂, a₃, \dots

be a G.P. of increasing positive numbers. If

a₃a₅ = 729

and

a₂ + a₄ = (111)/(4)

, then

24(a₁ + a₂ + a₃)

is equal to:

A. 131
B. 130
C. 129
D. 128

Solution

Related Formula

For a geometric sequence configuration with first term $a$ $a$ and common ratio $r$ $r$:

aₙ = a \cdot rⁿ⁻¹

Core Logic

Convert information markers using parameter notations:

a₃ a₅ = (ar²)(ar⁴) = a² r⁶ = 729 ar³ = 27

From second expression block data:

a₂ + a₄ = ar + ar³ = (111)/(4)

Substitute $ar³ = 27$ $ar^3 = 27$ directly into the linear equation block:

ar + 27 = (111)/(4) ar = (111)/(4) - 27 = (3)/(4)

Step 1: Finding parameters a and r

Divide the calculated components to evaluate the ratio:

(ar³)/(ar) = (27)/(3/4) r² = 36 r = 6

(Choose $+6$ $+6$ because terms must stay strictly positive ).

Find value for first base variable $a$ $a$:

a(6) = (3)/(4) a = (1)/(8)

Step 2: Sum configuration resolving

Now compute targeted expansion expression value:

24(a₁ + a₂ + a₃) = 24(a + ar + ar²) = 24a(1 + r + r²)

= 24 \cdot ((1)/(8)) \cdot (1 + 6 + 36) = 3 \cdot 43 = 129

Pattern Recognition

Product entries like $a₃ a₅ = a₄²$ $a_3 a_5 = a_4^2$ help identify the central term index value quickly in symmetric geometric progressions.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q62 jee_main_2025_04_april_evening Telescoping Series

If the sum of the first 20 terms of the series

4 . 14 + 3 . 1 ^ 2 + 1 ^ 4 + 4 . 24 + 3 . 2 ^ 2 + 2 ^ 4 + 4 . 34 + 3 . 3 ^ 2 + 3 ^ 4 + 4 . 44 + 3 . 4 ^ 2 + 4 ^ 4 +

is

mn

, where m and n are coprime, then

m + n

is equal to:

A. $423$ $423$
B. $420$ $420$
C. $421$ $421$
D. $422$ $422$

Solution

Core Logic

The general term $Tᵣ$ $T_r$ of the series can be written as:

Tᵣ = (4r)/(r⁴ + 3r² + 4)

Let's factorize the denominator by completing the square metric:

r⁴ + 3r² + 4 = (r⁴ + 4r² + 4) - r² = (r² + 2)² - r²

Using the difference of squares identity $A² - B² = (A-B)(A+B)$ $A^2 - B^2 = (A-B)(A+B)$:

r⁴ + 3r² + 4 = (r² - r + 2)(r² + r + 2)

Step 1: Partial Fraction Decomposition

Express $Tᵣ$ $T_r$ using partial fractions split:

Tᵣ = (4r)/((r² - r + 2)(r² + r + 2)) = 2 [ (1)/(r² - r + 2) - (1)/(r² + r + 2) ]

Notice that if we define $V(r) = r² - r + 2$ $V(r) = r^2 - r + 2$, then $V(r+1) = (r+1)² - (r+1) + 2 = r² + 2r + 1 - r - 1 + 2 = r² + r + 2$ .

Thus, $Tᵣ = 2[V(r) - V(r+1)]$ $T_r = 2\big[V(r) - V(r+1)\big]$, which sets up a clear telescoping sum formulation.

Step 2: Evaluating the Sum of 20 Terms

Summing from $r = 1$ $r = 1$ to $20$ $20$:

S₂₀ = Σr=1²⁰ Tᵣ = 2 Σr=1²⁰ [ (1)/(r² - r + 2) - (1)/(r² + r + 2) ]

= 2 [ ((1)/(2) - (1)/(4)) + ((1)/(4) - (1)/(8)) + + ((1)/(20² - 20 + 2) - (1)/(20² + 20 + 2)) ]

All sequential middle terms cancel completely, leaving only first and final values:

S₂₀ = 2 [ (1)/(2) - (1)/(422) ] = 1 - (1)/(211) = (210)/(211)

Since $210$ $210$ and $211$ $211$ are coprime, $m = 210$ $m = 210$ and $n = 211$ $n = 211$.

Step 3: Calculating m + n

Combining both values:

m + n = 210 + 211 = 421

Pattern Recognition

The polynomial factorization $r⁴ + a²r² + b⁴$ $r^4 + a^2r^2 + b^4$ is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence.

Chapter Mix

Class 11 Mathematics: Sequences and Series

The number of common terms in the progressions 4, 9, 14, 19,$4, 9, 14, 19, \dots$ up to 25th$25^{th}$ term and 3, 6, 9, 12,$3, 6, 9, 12, \dots$ up to 37th$37^{th}$ term is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Forming the Common AP

Step 2: Bounding the Sequence

Pattern Recognition

Chapter Mix

Reference Study Guides

More Sequences and Series Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Evaluating Sequence Parameters

Step 2: Summing the Target Range

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Parameters and Indices

Step 2: Computing the Target Segment Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Summing the Series

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding parameters a and r

Step 2: Sum configuration resolving

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Partial Fraction Decomposition

Step 2: Evaluating the Sum of 20 Terms

Step 3: Calculating m + n

Pattern Recognition

Chapter Mix

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The number of common terms in the progressions $4, 9, 14, 19,$ $4, 9, 14, 19, \dots$ up to $25th$ $25^{th}$ term and $3, 6, 9, 12,$ $3, 6, 9, 12, \dots$ up to $37th$ $37^{th}$ term is :