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Sequences and Series appeared 55 times across 3 years — 6.4% of Mathematics. This question is from Special Series.

Year 2026 2025 2024 Total
Questions 17 24 14 55

Let α = 1² + 4² + 8² + 13² + 19² + 26² + upto 10 terms and β = Σn=1¹⁰ n⁴. If 4α - β = 55k + 40, then k is equal to

Numerical Answer Type:
Enter a numerical value Answer: 353 to 353 +4 marks

Solution & Explanation

Related Formula
Method of differences for a sequence: Vₙ - Vₙ₋₁ = Tₙ
Core Logic

The base terms inside the squares form a sequence: 1, 4, 8, 13, 19, 26 The differences between consecutive terms are: 3, 4, 5, 6, 7 Since the first differences are an Arithmetic Progression, the general term of the inner sequence is a quadratic in n: Tₙ = an² + bn + c. Using n=1: 1 = a + b + c Using n=2: 4 = 4a + 2b + c Using n=3: 8 = 9a + 3b + c Solving this system: (4a + 2b + c) - (a + b + c) = 3 ⇒ 3a + b = 3 (9a + 3b + c) - (4a + 2b + c) = 4 ⇒ 5a + b = 4 Subtracting these gives: 2a = 1 ⇒ a = 1/2. Then 3(1/2) + b = 3 ⇒ b = 3/2. Finally 1/2 + 3/2 + c = 1 ⇒ c = -1. Inner sequence Tₙ = (1)/(2)n² + (3)/(2)n - 1.

Step 1: Calculating alpha structure

The series is α = Σn=1¹⁰ (Tₙ)².

4α = Σn=1¹⁰ 4((n² + 3n - 2)/(2))² = Σn=1¹⁰ (n² + 3n - 2)²

Expand the squared trinomial:

(n² + 3n - 2)² = n⁴ + 9n² + 4 + 6n³ - 4n² - 12n = n⁴ + 6n³ + 5n² - 12n + 4
Step 2: Applying given target relation

We are given β = Σn=1¹⁰ n⁴. So, 4α - β = Σn=1¹⁰ (n⁴ + 6n³ + 5n² - 12n + 4) - Σn=1¹⁰ n⁴

4α - β = Σn=1¹⁰ (6n³ + 5n² - 12n + 4)
Step 3: Calculating summation limits

Evaluate each standard summation up to n=10: Σ n³ = (10 × 11 / 2)² = 55² = 3025 Σ n² = (10 × 11 × 21) / 6 = 385 Σ n = (10 × 11) / 2 = 55 Σ 4 = 40

4α - β = 6(3025) + 5(385) - 12(55) + 40 = 18150 + 1925 - 660 + 40 = 19455

We are given 4α - β = 55k + 40.

19455 = 55k + 40

19415 = 55k

k = (19415)/(55) = 353
Pattern Recognition

Recognizing arithmetic progressions in the first-order differences immediately specifies a quadratic general term An²+Bn+C. Expanding and cancelling highest-order summation terms drastically simplifies standard sums.

Chapter Mix

Class 11 Maths: Sequences and Series

Reference Study Guides

More Sequences and Series Previous-Year Questions — Page 5

Q70 jee_main_2025_03_april_evening Special Series
The sum 1 + (1+3)/(2!) + (1+3+5)/(3!) + (1+3+5+7)/(4!) + is equal to
  • A. 6e
  • B. 4e
  • C. 3e
  • D. 2e

Solution

Related Formula

Sum of first r odd natural numbers:

Σk=1r (2k-1) = r²

Exponential series expansion:

Σr=0∞ (1)/(r!) = e
Core Logic

Let's find the general r-th term of the series:

Tᵣ = (1 + 3 + 5 + + (2r-1))/(r!) = (r²)/(r!) = (r)/((r-1)!)
Step 1: Expressing term in terms of sum limits

Let's write r = (r-1) + 1:

Tᵣ = (r-1+1)/((r-1)!) = (1)/((r-2)!) + (1)/((r-1)!)

Our infinite sum is:

S = Σr=1∞ Tᵣ = Σr=2∞ (1)/((r-2)!) + Σr=1∞ (1)/((r-1)!)

Both sums are standard representations of the exponential expansion.

Step 2: Summing the parts
  • First part: Σr=2∞ (1)/((r-2)!) = 1 + (1)/(1!) + (1)/(2!) + = e
  • Second part: Σr=1∞ (1)/((r-1)!) = 1 + (1)/(1!) + (1)/(2!) + = e
Total Sum S = e + e = 2e
Pattern Recognition

The general term containing r² in summation with factorials converges to 2e. Remember the shortcut: Σ (r²)/(r!) = 2e, Σ (r³)/(r!) = 5e. It is extremely useful to memorize these common limits.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability

Q63 jee_main_2025_07_april_morning Arithmetico-Geometric Progression
Let x₁, x₂, x₃, x₄ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x₁, x₂, x₃, x₄ then the resulting numbers are in an arithmetic progression. Then the value of (1)/(24) (x₁ x₂ x₃ x₄) is:
  • A. 72
  • B. 18
  • C. 36
  • D. 216

Solution

Related Formula

For a geometric progression, the terms can be set as a, ar, ar², ar³. For three terms A, B, C to be in arithmetic progression, they must satisfy: 2B = A + C

Core Logic

Let the elements be x₁ = a, x₂ = ar, x₃ = ar², x₄ = ar³. After the specified subtractions, the sequence becomes:

a - 2, ar - 7, ar² - 9, ar³ - 5

Since this sequence is in AP, we form two separate common difference linear linkages:

2(ar - 7) = (a - 2) + (ar² - 9) 2ar - 14 = ar² + a - 11 ar² - 2ar + a + 3 = 0 (1) 2(ar² - 9) = (ar - 7) + (ar³ - 5) 2ar² - 18 = ar³ + ar - 12 ar³ - 2ar² + ar + 6 = 0 (2)
Step 1: Solve the Simultaneous Polynomials

Multiply equation (1) by r:

ar³ - 2ar² + ar + 3r = 0 (3)

Subtract equation (3) from equation (2):

(ar³ - 2ar² + ar + 6) - (ar³ - 2ar² + ar + 3r) = 0 6 - 3r = 0 3r = 6 r = 2

Substitute r = 2 back into equation (1):

a(2)² - 2a(2) + a + 3 = 0 4a - 4a + a + 3 = 0 a = -3
Step 2: Find the Continuous Product Value

The continuous product term is:

x₁x₂x₃x₄ = a · ar · ar² · ar³ = a⁴ r⁶ x₁x₂x₃x₄ = (-3)⁴ · (2)⁶ = 81 × 64 = 5184

Now divide by 24 as required:

(1)/(24)(5184) = 216
Pattern Recognition

Notice that multiplying the first AP condition equation by r perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q jee_main_2025_08_april_evening Infinite Series
If (1)/(1⁴) + (1)/(2⁴) + (1)/(3⁴) + ∞ = (π⁴)/(90), and (1)/(1⁴) + (1)/(3⁴) + (1)/(5⁴) + ∞ = α (1)/(2⁴) + (1)/(4⁴) + (1)/(6⁴) + ∞ = β then (α)/(β) is equal to
  • A. 23
  • B. 18
  • C. 15
  • D. 14

Solution

Related Formula
Total Sum = α + β
Core Logic

Factor out common fractions from the even terms component (β) to represent it as a scalar multiple of the universal sum sequence.

Step 1: Simplify the Even Terms Series
β = (1)/(2⁴) + (1)/(4⁴) + (1)/(6⁴) + = (1)/(2⁴) ( (1)/(1⁴) + (1)/(2⁴) + (1)/(3⁴) + ) β = (1)/(16) ( (π⁴)/(90) )
Step 2: Express Alpha by Remainder Deduction

Since total sum equals α + β:

α = Total Sum - β = (π⁴)/(90) - (1)/(16) ( (π⁴)/(90) ) = (15)/(16) ( (π⁴)/(90) )
Step 3: Compute the Relative Ratio

(α)/(β) = ((15)/(16) ( (π⁴)/(90) ))/((1)/(16) ( (π⁴)/(90) )) = 15

Pattern Recognition

For alternating p-series powers like Σ n-p, the even component fractions always condense via factor steps to 2-p · Stotal, decoupling power values cleanly from final simple scalar quotients.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q57 jee_main_2025_29_jan_evening Infinite Geometric Progression
Let S = N 0. Define a relation \mathbf{R} from S to R by: R = (x, y): ₑ y = x ₑ ((2)/(5)), x in S, y in R . Then, the sum of all the elements in the range of R is equal to
  • A. (3)/(2)
  • B. (5)/(3)
  • C. (10)/(9)
  • D. (5)/(2)

Solution

Related Formula

Sum of an infinite geometric progression with |r| < 1:

S∞ = (a)/(1 - r)
Core Logic

From the definition of the relation:

ₑ y = x ₑ((2)/(5)) ₑ y = ₑ((2)/(5))^x y = ((2)/(5))^x

Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening
Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening

Since x in S = 0, 1, 2, 3,, the output values of y represent elements of the range.

Step 1: Compute Infinite Sum

Generating elements by plugging in values of x: For x = 0 y = 1 For x = 1 y = (2)/(5) For x = 2 y = ((2)/(5))²

Sum of elements in the range:

Sum = 1 + ((2)/(5))¹ + ((2)/(5))² + = (1)/(1 - (2)/(5)) = (5)/(3)
Pattern Recognition

Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario.

Chapter Mix

Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions

Q73 jee_main_2025_29_jan_evening Arithmetic Progression Properties
Let a₁, a₂, …, a₂₀₂₄ be an Arithmetic Progression such that a₁ + (a₅ + a₁₀ + a₁₅ + … + a₂₀₂₀) + a₂₀₂₄ = 2233. Then a₁ + a₂ + a₃ + … + a₂₀₂₄ is equal to
Numerical Answer. Answer: 11132 to 11132

Solution

Related Formula

Symmetry identity rule inside Arithmetic Progressions:

ak + an-k+1 = a₁ + aₙ
Core Logic

Group matching paired steps equidistant from sequence boundary ends:

a₁ + a₂₀₂₄ = a₅ + a₂₀₂₀ = a₁₀ + a₂₀₁₅ =

The sequence of inner indices follows an AP tracking loop:

5, 10, 15, , 2020

Calculate internal block element count N:

2020 = 5 + (N-1)5 2015 = 5(N-1) N - 1 = 403 N = 404 terms
Step 1: Simplify Expression Equations

Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding a₁ and a₂₀₂₄ introduces one more pair, resulting in 203 identical sum blocks:

203(a₁ + a₂₀₂₄) = 2233 a₁ + a₂₀₂₄ = (2233)/(203) = 11
Step 2: Evaluate the Total Sum

Using the standard AP sum formula:

S₂₀₂₄ = (2024)/(2)(a₁ + a₂₀₂₄) = 1012 × 11 = 11132
Pattern Recognition

Progressions possess natural positional balance. Grouping symmetrical index pairs (ak + an-k+1) allows factoring out variable steps right away.

Chapter Mix

Class 11 Mathematics: Sequences and Series

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