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

Year 2026 2025 2024 Total
Questions 17 24 14 55

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:

Solution & Explanation

Core Logic

The general term Tᵣ 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):

r⁴ + 3r² + 4 = (r² - r + 2)(r² + r + 2)
Step 1: Partial Fraction Decomposition

Express Tᵣ 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, 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)], which sets up a clear telescoping sum formulation.

Step 2: Evaluating the Sum of 20 Terms

Summing from r = 1 to 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 and 211 are coprime, m = 210 and n = 211.

Step 3: Calculating m + n

Combining both values:

m + n = 210 + 211 = 421
Pattern Recognition

The polynomial factorization r⁴ + a²r² + b⁴ 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

Reference Study Guides

More Sequences and Series Previous-Year Questions — Page 7

Q65 jee_main_2025_04_april_evening Arithmetic Progression
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that D = d + 3, ~d > 0. If p + qp - q = (19)/(5), then p - q is equal to
  • A. 600
  • B. 450
  • C. 630
  • D. 540

Solution

Core Logic

Let the 3 elements of set A in A.P. be a-d, a, a+d. Their sum is 3a = 36 a = 12. Their product is p = a(a² - d²) = 12(144 - d²).

Similarly, let the 3 elements of set B be b-D, b, b+D. Their sum is 3b = 36 b = 12. Their product is q = b(b² - D²) = 12(144 - D²).

Step 1: Using the Ratio Condition

We are given the relation:

(p + q)/(p - q) = (19)/(5)

Using componendo and dividendo:

(p)/(q) = (19 + 5)/(19 - 5) = (24)/(14) = (12)/(7)

Substitute the expression blocks for p and q:

(12(144 - d²))/(12(144 - D²)) = (12)/(7) (144 - d²)/(144 - D²) = (12)/(7) 7(144 - d²) = 12(144 - D²)
Step 2: Substituting D in terms of d

We are given D = d + 3:

7(144 - d²) = 12(144 - (d + 3)²) 1008 - 7d² = 12(144 - (d² + 6d + 9)) 1008 - 7d² = 12(135 - d² - 6d) = 1620 - 12d² - 72d 5d² + 72d - 612 = 0

Solving this quadratic equation:

(d - 6)(5d + 102) = 0

Since d > 0, we choose d = 6. This implies D = 6 + 3 = 9.

Step 3: Finding p - q

Now calculate the targeted metric:

p - q = 12(144 - d²) - 12(144 - D²) = 12(D² - d²) p - q = 12(9² - 6²) = 12(81 - 36) = 12(45) = 540
Pattern Recognition

For 3-element symmetric AP sequences, choosing terms as x-d, x, x+d ensures the sum isolates the middle term instantly (3x = S). This drastically drops algebraic variables from the start.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q53 jee_main_2025_04_april_morning Arithmetic Progression
Let A = 1, 6, 11, 16, and B = 9, 16, 23, 30, be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A B) is
  • A. 3814
  • B. 4027
  • C. 3761
  • D. 4003

Solution

Related Formula

Set Principle of Inclusion-Exclusion:

n(A B) = n(A) + n(B) - n(A B)
Core Logic

Find the last terms of both progressions: For set A: a₁ = 1, d₁ = 5 T₂₀₂₅ = 1 + (2025 - 1) × 5 = 10121. For set B: b₁ = 9, d₂ = 7 T₂₀₂₅ = 9 + (2025 - 1) × 7 = 14177.

The intersection set A B forms an AP with a common difference d = LCM(5, 7) = 35. The first common term is 16.

Step 1: Find Common Terms Count

The general term of the common AP must satisfy:

Tₙ = 16 + (n - 1) × 35 ≤ (10121, 14177) = 10121 (n - 1) × 35 ≤ 10105 n - 1 ≤ 288.71 n = 289
Step 2: Total Distinct Terms

Apply the inclusion-exclusion principle:

n(A B) = 2025 + 2025 - 289 = 3761
Pattern Recognition

Common terms of two APs always generate a new AP whose common difference is the LCM of the individual common differences. Always verify the upper limit bound using the smaller of the two final values.

Chapter Mix

Class 11 Mathematics: Sequence and Series

Q61 jee_main_2025_04_april_morning Special Series
1 + 3 + 5² + 7 + 9² + upto 40 terms is equal to
  • A. 43890
  • B. 41880
  • C. 33980
  • D. 40870

Solution

Related Formula

Summation Identities:

Σ r = (n(n+1))/(2), Σ r² = (n(n+1)(2n+1))/(6)
Core Logic

Split the 40-term series into two sub-series of 20 terms each: Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd): 1² + 5² + 9² + upto 20 terms. General term Tᵣ = (4r - 3)². Series 2 (linear terms at positions 2, 4, 6...): 3 + 7 + 11 + upto 20 terms. General term tᵣ = (4r - 1).

Step 1: Formulate Total Sigma Expression
Sum = Σr=1²⁰ [ (4r - 3)² + (4r - 1) ] Sum = Σr=1²⁰ (16r² - 24r + 9 + 4r - 1) = Σr=1²⁰ (16r² - 20r + 8) Sum = 16Σr=1²⁰ r² - 20Σr=1²⁰ r + 8Σr=1²⁰ 1
Step 2: Arithmetic Evaluation
Σr=1²⁰ r² = (20 × 21 × 41)/(6) = 2870 Σr=1²⁰ r = (20 × 21)/(2) = 210 Sum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880
Pattern Recognition

When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials.

Chapter Mix

Class 11 Mathematics: Sequence and Series

Q60 jee_main_2025_07_april_evening Arithmetic Progression
Let aₙ be the nth term of an A. P. If Sₙ = a₁ + a₂ + a₃ + + aₙ = 700, a₆ = 7 and S₇ = 7, then aₙ is equal to:
  • A. 56
  • B. 65
  • C. 64
  • D. 70

Solution

Related Formula

Sum of first n terms of an AP is given by:

Sₙ = (n)/(2)[2a + (n-1)d]
Core Logic

Given specifications:

  • a₆ = 7 a + 5d = 7 (ii)
  • S₇ = 7 (7)/(2)(2a + 6d) = 7 a + 3d = 1 (iii)
  • Subtracting (iii) from (ii):

2d = 6 d = 3

Substituting d=3 into (iii):

a + 3(3) = 1 a = -8
Step 1: Find n from Sn = 700

Substitute a = -8 and d = 3 into the equation for Sₙ = 700:

700 = (n)/(2)[2(-8) + (n-1)3] 1400 = n[-16 + 3n - 3] 3n² - 19n - 1400 = 0

Factoring the quadratic equation:

(3n + 56)(n - 25) = 0

Since n must be a positive integer, n = 25.

Step 2: Determine standard term value

We need to find a₂₅ corresponding to index n=25:

a₂₅ = a + 24d a₂₅ = -8 + 24(3) = -8 + 72 = 64
Pattern Recognition

When Sₙ and specific terms are given, prioritize finding the first term a and common difference d through simple elimination headers before targeting the value of n.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q70 jee_main_2025_07_april_evening Geometric Progression
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :
  • A. 745
  • B. 755
  • C. 750
  • D. 757

Solution

Related Formula

Sum of first n terms of a Geometric Progression (GP) is:

Sₙ = (a(rⁿ - 1))/(r - 1)
Core Logic

Let the first term be a and common ratio be r. Given:

  • ar + ar³ + ar⁵ = 21 ar(1 + r² + r⁴) = 21 (1)
  • ar⁷ + ar⁹ + ar¹¹ = 15309 ar⁷(1 + r² + r⁴) = 15309 (2)
  • Dividing equation (2) by equation (1):

(ar⁷)/(ar) = (15309)/(21) r⁶ = 729 r = 3
Step 1: Solve for a

Substitute r = 3 into equation (1):

a(3)(1 + 9 + 81) = 21 3a(91) = 21 a = (7)/(91) = (1)/(13)
Step 2: Find Sum of 9 terms

Evaluating S₉:

S₉ = (a(r⁹ - 1))/(r - 1) = ((1)/(13)(3⁹ - 1))/(3 - 1) = (19683 - 1)/(26) = (19682)/(26) = 757
Pattern Recognition

Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio r^k instantly.

Chapter Mix

Class 11 Mathematics: Sequences and Series

More Sequences and Series Questions — jee_main_2025_04_april_evening

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