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

Year 2026 2025 2024 Total
Questions 17 24 14 55

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 :

Solution & Explanation

Related Formula
Tₙ = a + (n-1)d Dcommon = LCM(d₁, d₂)
Core Logic

First Progression (S₁): 4, 9, 14, 19, Common difference d₁ = 5. Last term (T₂₅) = 4 + (25-1)5 = 4 + 120 = 124.

Second Progression (S₂): 3, 6, 9, 12, Common difference d₂ = 3. Last term (T₃₇) = 3 + (37-1)3 = 3 + 108 = 111.

Step 1: Forming the Common AP

By inspecting the sequences, the first common term (acommon) is 9. The common difference of the new series is the LCM of the original differences:

Dcommon = LCM(5, 3) = 15

Thus, the common terms form a new AP: 9, 24, 39, 54,

Step 2: Bounding the Sequence

The last term of the common AP must be less than or equal to the smallest maximum limit of the two series. Here, (124, 111) = 111. So, the n-th term of the common sequence is bounded by 111:

9 + (n-1)15 ≤ 111

15(n-1) ≤ 102

(n-1) ≤ (102)/(15) = 6.8

n ≤ 7.8 Since n must be an integer, n = 7.

Pattern Recognition

The common terms of two APs always form a new AP. Its common difference is the LCM of the original differences. Find the first common term manually, then cap the n-th term inequality with the smallest end-boundary of the original sets.

Chapter Mix

Class 11 Maths: Sequences and Series

Reference Study Guides

More Sequences and Series Previous-Year Questions — Page 9

Q63 jee_main_2025_29_jan_morning Arithmetic Progression
Consider an A.P. of positive integers, whose \sum of the first three terms is 54 and the \sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:
  • A. 84
  • B. 122
  • C. 90
  • D. 108

Solution

Related Formula
Sₙ = (n)/(2) [2a + (n-1)d] aₙ = a + (n-1)d
Core Logic

Given S₃ = 54 3a + 3d = 54 a + d = 18. Express S₂₀ as:

S₂₀ = (20)/(2)[2a + 19d] = 10(2a + 19d)

Substitute a = 18 - d into the expression:

S₂₀ = 10[2(18 - d) + 19d] = 10(36 + 17d)
Step 1: Formulate Inequality and Constraint Bound

Given 1600 < S₂₀ < 1800:

1600 < 10(36 + 17d) < 1800 160 < 36 + 17d < 180

124 < 17d < 144

(124)/(17) < d < (144)/(17) 7.29 < d < 8.47
Step 2: Isolate Integer Term parameters

Since the sequence consists of positive integers, common difference d must be an integer d = 8. Then a = 18 - 8 = 10.

Step 3: Calculate the 11th Term
a₁₁ = a + 10d = 10 + 10(8) = 90
Pattern Recognition

Diophantine properties (integer conditions) drastically restrict valid inequality windows. Always check parameters for strict divisibility to skip unnecessary computation.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q17 jee_main_2024_01_february_morning Arithmetic and Geometric Progressions
Let 3, a, b, c be in A.P. and 3, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of a, b and c is:
  • A. -4
  • B. -1
  • C. 13
  • D. 11

Solution

Related Formula
  • An Arithmetic Progression (A.P.) with common difference d sets consecutive terms as: Tₙ = T₁ + (n-1)d
  • A Geometric Progression (G.P.) ensures: T₂² = T₁ · T₃
Core Logic

Since 3, a, b, c are elements of an A.P., let d denote the common difference:

  • a = 3 + d
  • b = 3 + 2d
  • c = 3 + 3d
  • Substituting these values into the sequence configurations of the given G.P. (3, a-1, b+1, c+9):

G.P. terms: 3, (3+d-1), (3+2d+1), (3+3d+9) G.P. terms: 3, 2+d, 4+2d, 12+3d
Step 1: Compute the Common Difference

Using the geometric mean property for the first three terms (3, 2+d, 4+2d):

(2+d)² = 3(4 + 2d) 4 + 4d + d² = 12 + 6d d² - 2d - 8 = 0 (d-4)(d+2) = 0 d = 4 or d = -2
Step 2: Evaluate both cases for the Progressions
  • Case A: If d = 4
  • The G.P. sequence reads: 3, 6, 12, 24 (common ratio r=2, valid layout). The values are: a = 7, b = 11, c = 15.

  • Case B: If d = -2
  • The G.P. sequence reads: 3, 0, 0, 6 (contains zeros, violating standard geometric definitions).

    Hence, select d = 4.

Step 3: Calculate the Final Arithmetic Mean

The required arithmetic mean of a, b, c is:

Arithmetic Mean = (a+b+c)/(3) = (7+11+15)/(3) = (33)/(3) = 11
Pattern Recognition

Sees: Transition parameters mapping from A.P. linear spacing into G.P. ratios. Shortcut: Notice that the arithmetic mean of a, b, c is exactly equal to the middle value b for any linear sequence. Thus, finding b = 3 + 2(4) = 11 directly yields the final answer.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q24 jee_main_2024_01_february_morning Common Terms of Two APs
Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11,..., 404 be two arithmetic progressions. Then the sum of the common terms in them is equal to
Numerical Answer. Answer: 6699 to 6699

Solution

Related Formula
  • General term of an AP: Tₙ = a + (n-1)d
  • Sum of n terms of an AP: Sₙ = (n)/(2)[2a + (n-1)d]
  • The common difference of a series of common terms between two APs is given by the least common multiple of their respective common differences:
dcommon = LCM(d₁, d₂)
Core Logic

Let's analyze both arithmetic progressions:

  • AP 1: 3, 7, 11, 15, , 403 First term a₁ = 3, common difference d₁ = 4.
  • AP 2: 2, 5, 8, 11, , 404 First term a₂ = 2, common difference d₂ = 3.
  • By observation, the first identical value appearing in both series is 11. Therefore, the new common AP has:

  • First term a = 11
  • Common difference d = LCM(4, 3) = 12
Step 1: Determine the Number of Common Terms

The last term Tₙ of the common AP cannot exceed the boundary upper limits of either individual series (i.e., ≤ 403):

Tₙ = 11 + (n-1)12 ≤ 403

12(n-1) ≤ 392

n-1 ≤ 32.66 n = 33
Step 2: Calculate the Series Sum

Using the AP summation formula for 33 terms:

S₃₃ = (33)/(2) [ 2(11) + (33-1)12 ] S₃₃ = (33)/(2) [ 22 + 32 × 12 ] S₃₃ = (33)/(2) [ 22 + 384 ] S₃₃ = (33)/(2) × 406 = 33 × 203 = 6699
Pattern Recognition

Sees: Overlapping arithmetic progression series elements. Shortcut: Once the first matching number and the LCM are calculated, the maximum term inequality a + (n-1)d ≤ (L₁, L₂) directly maps out the total number of terms cleanly.

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q9 jee_main_2024_29_january_evening Arithmetic Progression
If ₑa, ₑb, ₑc are in an A.P. and ₑa - ₑ2b, ₑ2b - ₑ3c, ₑ3c - ₑa are also in an A.P, then a:b:c is equal to
  • A. 9 : 6 : 4
  • B. 16 : 4 : 1
  • C. 25 : 10 : 4
  • D. 6 : 3 : 2

Solution

Related Formula

If x, y, z are in A.P., then 2y = x + z.

Core Logic

From the first sequence condition:

2 ₑ b = ₑ a + ₑ c ₑ b² = ₑ(ac) b² = ac (i)

From the second sequence condition, the components are ₑ((a)/(2b)), ₑ((2b)/(3c)), ₑ((3c)/(a)):

2 ₑ((2b)/(3c)) = ₑ((a)/(2b)) + ₑ((3c)/(a)) ((2b)/(3c))² = (a)/(2b) × (3c)/(a) = (3c)/(2b) (4b²)/(9c²) = (3c)/(2b) 8b³ = 27c³ (b)/(c) = (3)/(2) (ii)
Step 1: Finding Ratios

Substituting c = (2b)/(3) into equation (i):

b² = a ((2b)/(3)) b = (2a)/(3) (a)/(b) = (3)/(2)

Thus, consolidating all parts: a : b = 9 : 6 b : c = 6 : 4

a : b : c = 9 : 6 : 4
Pattern Recognition

Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2 x = x²).

Chapter Mix

Class 11 Mathematics: Sequences and Series

Q13 jee_main_2024_29_january_evening Geometric Progression
If each term of a geometric progression a₁, a₂, a₃, … with a₁ = (1)/(8) and a₂ ≠ a₁, is the arithmetic mean of the next two terms and Sₙ = a₁ + a₂ + … + aₙ, then S₂₀ - S₁₈ is equal to
  • A. 2¹⁵
  • B. -2¹⁸
  • C. 2¹⁸
  • D. -2¹⁵

Solution

Related Formula
2aₙ = aₙ₊₁ + aₙ₊₂
Core Logic

Let the terms of the Geometric Progression have a common ratio r. Substituting the geometric forms into the arithmetic mean relationship:

2(a rⁿ⁻¹) = a rⁿ + a rⁿ⁺¹

Dividing out non-zero fields a rⁿ⁻¹:

2 = r + r² r² + r - 2 = 0 (r + 2)(r - 1) = 0

Since a₂ ≠ a₁, we have r ≠ 1. Thus, the common ratio is r = -2.

Step 1: Evaluating the Target Partial Difference

We need to evaluate:

S₂₀ - S₁₈ = T₁₉ + T₂₀ T₁₉ + T₂₀ = a r¹⁸ + a r¹⁹ = a r¹⁸(1 + r)

Substituting a = (1)/(8) and r = -2:

T₁₉ + T₂₀ = (1)/(8) (-2)¹⁸ (1 - 2) = (1)/(2³) · 2¹⁸ · (-1) = -2¹⁵
Pattern Recognition

Partial sum differences simplify into standard standalone term values (Sₙ - Sₙ₋₂ = Tₙ + Tₙ₋₁). This eliminates the need to apply long fraction sum formats.

Chapter Mix

Class 11 Mathematics: Sequences and Series

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