Class 11 Notes

Class 11 Mathematics Notes Chapter 9 (Chapter 9) – Examplar Problems (English) Book

Philoid

Philoid

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Examplar Problems (English)
Alright class, let's get straight into Chapter 9: Sequences and Series from your NCERT Exemplar. This chapter is crucial not just for your Class 11 exams but forms the foundation for many quantitative aptitude sections in government exams. Pay close attention to the formulas and concepts.

Chapter 9: Sequences and Series - Detailed Notes

1. Sequence:

  • A sequence is an ordered list of numbers (called terms) according to some definite rule.
  • Notation: a₁, a₂, a₃, ..., aₙ, ... where aₙ is the nth term.
  • Finite Sequence: Contains a limited number of terms.
  • Infinite Sequence: Continues indefinitely.

2. Series:

  • A series is the sum of the terms of a sequence.
  • Notation: a₁ + a₂ + a₃ + ... + aₙ + ...
  • Sigma Notation (Σ): Used to represent a series compactly.
    • Σⁿ_{k=1} aₖ = a₁ + a₂ + ... + aₙ

3. Progression:

  • A sequence whose terms follow a certain pattern or rule. We primarily study Arithmetic Progression (AP), Geometric Progression (GP), and sometimes Harmonic Progression (HP).

4. Arithmetic Progression (AP):

  • Definition: A sequence where the difference between any term (except the first) and its preceding term is constant. This constant difference is called the common difference (d).
    • d = aₙ₊₁ - aₙ
  • General Term (nth term): If 'a' is the first term and 'd' is the common difference, the nth term is:
    • aₙ = a + (n-1)d
  • Sum of first n terms (Sₙ):
    • Sₙ = n/2 [2a + (n-1)d]
    • Sₙ = n/2 [a + l] (where l = aₙ is the last term)
  • Properties of AP:
    • If a constant is added/subtracted to each term, the resulting sequence is also an AP with the same common difference.
    • If each term is multiplied/divided by a non-zero constant k, the resulting sequence is also an AP with common difference kd or d/k respectively.
    • The sum of terms equidistant from the beginning and end is constant and equal to the sum of the first and last term (a₁ + aₙ = a₂ + aₙ₋₁ = ...).
    • Three terms in AP: (a-d), a, (a+d)
    • Four terms in AP: (a-3d), (a-d), (a+d), (a+3d)
    • Five terms in AP: (a-2d), (a-d), a, (a+d), (a+2d)
  • Arithmetic Mean (AM):
    • The AM of two numbers a and b is (a+b)/2.
    • If a, A, b are in AP, then A = (a+b)/2 is the AM.
    • To insert 'n' AMs between a and b: Let the AMs be A₁, A₂, ..., Aₙ. Then a, A₁, A₂, ..., Aₙ, b form an AP with (n+2) terms. The common difference d = (b-a)/(n+1). Then A₁ = a+d, A₂ = a+2d, ..., Aₙ = a+nd.

5. Geometric Progression (GP):

  • Definition: A sequence where the ratio of any term (except the first) to its preceding term is constant. This constant ratio is called the common ratio (r).
    • r = aₙ₊₁ / aₙ (r ≠ 0)
  • General Term (nth term): If 'a' is the first term and 'r' is the common ratio:
    • aₙ = arⁿ⁻¹
  • Sum of first n terms (Sₙ):
    • Sₙ = a(rⁿ - 1) / (r - 1) or a(1 - rⁿ) / (1 - r), if r ≠ 1
    • Sₙ = na, if r = 1
  • Sum of an Infinite GP (S∞):
    • If |r| < 1, the sum converges to: S∞ = a / (1 - r)
    • If |r| ≥ 1, the sum diverges (does not have a finite value).
  • Properties of GP:
    • If each term is multiplied/divided by a non-zero constant, the resulting sequence is also a GP with the same common ratio.
    • The reciprocals of the terms of a GP also form a GP.
    • If each term is raised to the same power k, the resulting sequence is also a GP with common ratio rᵏ.
    • The product of terms equidistant from the beginning and end is constant and equal to the product of the first and last term (a₁ * aₙ = a₂ * aₙ₋₁ = ...).
    • Three terms in GP: a/r, a, ar
    • Four terms in GP: a/r³, a/r, ar, ar³
    • Five terms in GP: a/r², a/r, a, ar, ar²
  • Geometric Mean (GM):
    • The GM of two positive numbers a and b is √(ab).
    • If a, G, b are in GP, then G = √(ab) is the GM (assuming a, b have the same sign).
    • To insert 'n' GMs between a and b: Let the GMs be G₁, G₂, ..., Gₙ. Then a, G₁, G₂, ..., Gₙ, b form a GP with (n+2) terms. The common ratio r = (b/a)^(1/(n+1)). Then G₁ = ar, G₂ = ar², ..., Gₙ = arⁿ.

6. Relationship between AM and GM:

  • For any two positive numbers a and b:
    • AM ≥ GM, i.e., (a+b)/2 ≥ √(ab)
    • Equality holds if and only if a = b.
  • This is extremely useful in problems involving maxima/minima.

7. Harmonic Progression (HP): (Less common, but good to know)

  • Definition: A sequence is in HP if the reciprocals of its terms form an AP.
    • Example: 1/2, 1/4, 1/6, ... is an HP because 2, 4, 6, ... is an AP.
  • nth term: To find the nth term of an HP, find the nth term of the corresponding AP and take its reciprocal. There's no direct formula for the sum of n terms of an HP.
  • Harmonic Mean (HM):
    • The HM of two non-zero numbers a and b is 2ab / (a+b).
    • If a, H, b are in HP, then H = 2ab / (a+b).
  • Relationship: For positive numbers, AM ≥ GM ≥ HM.

8. Sum of Special Series:

  • Sum of first n natural numbers:
    • Σn = 1 + 2 + ... + n = n(n+1)/2
  • Sum of squares of first n natural numbers:
    • Σn² = 1² + 2² + ... + n² = n(n+1)(2n+1)/6
  • Sum of cubes of first n natural numbers:
    • Σn³ = 1³ + 2³ + ... + n³ = [n(n+1)/2]² = (Σn)²

9. Arithmetico-Geometric Progression (AGP):

  • A series where each term is the product of the corresponding terms of an AP and a GP.
    • Form: a, (a+d)r, (a+2d)r², (a+3d)r³, ...
  • Method to find the sum (Sₙ):
    1. Write down the series Sₙ.
    2. Multiply Sₙ by the common ratio 'r' of the GP part (rSₙ).
    3. Subtract the second equation from the first (Sₙ - rSₙ).
    4. The resulting series will usually be a GP (or have a GP part), which can be summed easily.
    5. Solve for Sₙ.

10. Method of Differences:

  • Used when the differences between consecutive terms (a₂-a₁, a₃-a₂, ...) are in AP or GP.
  • Finding the nth term (aₙ):
    • Let Sₙ = a₁ + a₂ + ... + aₙ.
    • Write Sₙ again, shifted by one term: Sₙ = a₁ + a₂ + ... + aₙ₋₁ + aₙ.
    • Subtract: 0 = a₁ + (a₂-a₁) + (a₃-a₂) + ... + (aₙ-aₙ₋₁) - aₙ.
    • aₙ = a₁ + [(a₂-a₁) + (a₃-a₂) + ... + (aₙ-aₙ₋₁)]
    • The terms in the bracket form a new series (usually AP or GP) with (n-1) terms. Sum this series to find aₙ.
  • Finding the Sum (Sₙ): Once aₙ is found (often in terms of n), Sₙ = Σaₙ. Use the formulas for Σn, Σn², Σn³.

Multiple Choice Questions (MCQs)

  1. If the sum of n terms of an AP is given by Sₙ = 3n + 2n², then the common difference of the AP is:
    (A) 3
    (B) 2
    (C) 6
    (D) 4

  2. The third term of a GP is 4. The product of the first five terms is:
    (A) 4³
    (B) 4⁵
    (C) 4⁴
    (D) Cannot be determined

  3. If x, 2y, 3z are in AP and x, y, z are in GP, then the common ratio of the GP is:
    (A) 3
    (B) 1/3
    (C) 2
    (D) 1/2

  4. The sum of the infinite series 1 + 2/3 + 3/3² + 4/3³ + ... is:
    (A) 3
    (B) 9/4
    (C) 4/9
    (D) 9/2

  5. If the AM between two positive numbers a and b (a > b) is twice their GM, then a : b is:
    (A) (2 + √3) : (2 - √3)
    (B) (√2 + 1) : (√2 - 1)
    (C) (√3 + 2) : (√3 - 2)
    (D) (7 + 4√3) : 1

  6. The sum of the series 1² + 3² + 5² + ... up to n terms is:
    (A) n(2n-1)(2n+1)/3
    (B) n(n+1)(2n+1)/6
    (C) [n(n+1)/2]²
    (D) n(n+1)(n+2)/6

  7. If the pth, qth, and rth terms of an AP are a, b, c respectively, then a(q-r) + b(r-p) + c(p-q) equals:
    (A) 1
    (B) a+b+c
    (C) 0
    (D) pqr

  8. Let Sₙ denote the sum of the first n terms of an AP. If S₂ₙ = 3Sₙ, then S₃ₙ : Sₙ is equal to:
    (A) 4
    (B) 6
    (C) 8
    (D) 10

  9. The value of 2¹/⁴ ⋅ 4¹/⁸ ⋅ 8¹/¹⁶ ⋅ 16¹/³² ⋅ ... is:
    (A) 1
    (B) 2
    (C) 3/2
    (D) 4

  10. If 1/(a+b), 1/(2b), 1/(b+c) are in AP (a, b, c > 0), then a, b, c are in:
    (A) AP
    (B) GP
    (C) HP
    (D) None of these

Answer Key for MCQs:

  1. (D) 4 [Hint: a₁ = S₁ = 5, a₂ = S₂ - S₁ = (3(2)+2(2²)) - 5 = 14 - 5 = 9. d = a₂ - a₁ = 9 - 5 = 4]
  2. (B) 4⁵ [Hint: Let terms be a/r², a/r, a, ar, ar². Third term is a = 4. Product is a⁵ = 4⁵]
  3. (B) 1/3 [Hint: 2(2y) = x + 3z (AP). y² = xz (GP). Substitute x=y²/z in AP eq: 4y = y²/z + 3z => 4yz = y² + 3z². Let common ratio be r. x=a, y=ar, z=ar². Substitute: 4(ar)(ar²) = (ar)² + 3(ar²)². Simplify to get 4ar³ = a²r² + 3a²r⁴. Divide by ar² (assuming a,r≠0): 4r = a + 3ar². Also from AP: 4ar = a + 3ar². Comparing gives a=1, r=1/3 or solve the equations.]
  4. (B) 9/4 [Hint: This is an AGP. Let S = 1 + 2/3 + 3/3² + ... Multiply by 1/3. Subtract. S - S/3 = 1 + 1/3 + 1/3² + ... (Infinite GP). 2S/3 = 1/(1-1/3) = 3/2. S = 9/4]
  5. (A) (2 + √3) : (2 - √3) [Hint: (a+b)/2 = 2√(ab). Let a = kb. (kb+b)/2 = 2√(kb²). b(k+1)/2 = 2b√k. (k+1)/2 = 2√k. k+1 = 4√k. Squaring: (k+1)² = 16k => k²+2k+1=16k => k²-14k+1=0. Solve quadratic for k: k = [14 ± √(196-4)]/2 = 7 ± 4√3. Since a>b, k>1. So k = 7+4√3. a/b = 7+4√3 = (2+√3)² / 1 = (2+√3) / (2-√3) after rationalizing or recognizing (2+√3)² = 4+3+4√3 = 7+4√3. So a:b = (7+4√3):1 or (2+√3):(2-√3)]
  6. (A) n(2n-1)(2n+1)/3 [Hint: The nth term is (2n-1)². Sₙ = Σ(2n-1)² = Σ(4n² - 4n + 1) = 4Σn² - 4Σn + Σ1 = 4[n(n+1)(2n+1)/6] - 4[n(n+1)/2] + n. Simplify this expression.]
  7. (C) 0 [Hint: Let the first term be A and common difference be D. a = A+(p-1)D, b = A+(q-1)D, c = A+(r-1)D. Substitute these into the expression a(q-r) + b(r-p) + c(p-q). The terms with A cancel, and the terms with D cancel.]
  8. (B) 6 [Hint: Sₙ = n/2[2a+(n-1)d]. S₂ₙ = (2n)/2[2a+(2n-1)d]. S₃ₙ = (3n)/2[2a+(3n-1)d]. Given S₂ₙ = 3Sₙ => n[2a+(2n-1)d] = 3 * n/2[2a+(n-1)d]. Simplify to find relation between a and d (often a=d/2). Then find S₃ₙ / Sₙ = { (3n)/2[2a+(3n-1)d] } / { n/2[2a+(n-1)d] } = 3[2a+(3n-1)d] / [2a+(n-1)d]. Substitute relation between a and d. It simplifies to 6.]
  9. (B) 2 [Hint: Expression = 2¹/⁴ ⋅ (2²)¹/⁸ ⋅ (2³)¹/¹⁶ ⋅ (2⁴)¹/³² ⋅ ... = 2¹/⁴ ⋅ 2²/⁸ ⋅ 2³/¹⁶ ⋅ 2⁴/³² ⋅ ... = 2^(1/4 + 2/8 + 3/16 + 4/32 + ...). Let S = 1/4 + 2/8 + 3/16 + 4/32 + ... This is an AGP with a=1, d=1 and r=1/2 (starting from first term as a/r^0 = 1/2^1). Or view it as S = 1/2² + 2/2³ + 3/2⁴ + ... Multiply by 1/2: S/2 = 1/2³ + 2/2⁴ + ... Subtract: S - S/2 = 1/2² + 1/2³ + 1/2⁴ + ... (Infinite GP). S/2 = (1/4) / (1 - 1/2) = (1/4) / (1/2) = 1/2. So S = 1. The original expression is 2^S = 2¹ = 2.]
  10. (B) GP [Hint: If 1/(a+b), 1/(2b), 1/(b+c) are in AP, then 2/(2b) = 1/(a+b) + 1/(b+c). => 1/b = (b+c + a+b) / [(a+b)(b+c)]. => (a+b)(b+c) = b(a+2b+c). => ab + ac + b² + bc = ab + 2b² + bc. => ac + b² = 2b². => ac = b². This is the condition for a, b, c to be in GP.]

Make sure you understand the derivation of formulas and the conditions under which they apply. Practice problems from the Exemplar book thoroughly. Good luck!

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