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Algebra - Sequences and Series (Arithmetic and Geometric Progressions)

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Arithmetic Progression (AP): A sequence where the difference between any two consecutive terms is constant, known as the common difference dd. Visually, if you plot the terms of an AP against their position nn, the points lie on a straight line, representing a constant rate of change similar to a linear graph.

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Arithmetic Mean (AM): The arithmetic mean of two numbers aa and bb is the value a+b2\frac{a + b}{2}. If three numbers are in AP, the middle term is the AM of the other two. On a number line, the AM represents the exact midpoint between two values.

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Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio rr. In a visual representation, a GP with r>1r > 1 shows exponential growth (a curve sloping upwards), while a GP with 0<r<10 < r < 1 shows exponential decay (a curve flattening towards the x-axis).

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Geometric Mean (GM): For two positive numbers aa and bb, the GM is ab\sqrt{ab}. If a,G,ba, G, b are in GP, GG is the geometric mean. Geometrically, if aa and bb are the sides of a rectangle, the GM is the side of a square with the same area.

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Sum to Infinity: An infinite geometric series has a finite sum only if the absolute value of the common ratio is less than 1 (∣r∣<1|r| < 1). This concept can be visualized as an object moving half the remaining distance to a goal with every step; it gets closer and closer but represents a finite total displacement.

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Property of AM and GM: For any two positive real numbers aa and bb, the Arithmetic Mean is always greater than or equal to the Geometric Mean (AM≥GMAM \ge GM). They are equal only if a=ba = b.

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General Term and Sum: The general term (nthn^{th} term) allows for finding any specific value in a sequence without listing all previous terms. The sum of nn terms (series) represents the accumulation of values, which in an AP is the area of a trapezoidal shape formed by the discrete steps of the sequence.

📐Formulae

General term of an AP: an=a+(n−1)da_n = a + (n - 1)d

Sum of nn terms of an AP: Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n - 1)d] or Sn=n2(a+l)S_n = \frac{n}{2}(a + l), where ll is the last term

Arithmetic Mean (AM) between aa and bb: A=a+b2A = \frac{a + b}{2}

General term of a GP: an=arn−1a_n = ar^{n-1}

Sum of nn terms of a GP: Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} for r≠1r \neq 1

Sum to infinity of a GP: S∞=a1−rS_{\infty} = \frac{a}{1 - r}, provided ∣r∣<1|r| < 1

Geometric Mean (GM) between aa and bb: G=abG = \sqrt{ab}

💡Examples

Problem 1:

Find the 15th15^{th} term and the sum of the first 15 terms of the AP: 3,7,11,15,…3, 7, 11, 15, \dots

Solution:

  1. Identify the first term a=3a = 3 and the common difference d=7−3=4d = 7 - 3 = 4.
  2. Use the general term formula: a15=a+(15−1)d=3+14(4)=3+56=59a_{15} = a + (15 - 1)d = 3 + 14(4) = 3 + 56 = 59.
  3. Use the sum formula: S15=n2(a+l)=152(3+59)=152(62)=15×31=465S_{15} = \frac{n}{2}(a + l) = \frac{15}{2}(3 + 59) = \frac{15}{2}(62) = 15 \times 31 = 465.

Explanation:

To solve for any component of an AP, first determine the starting value aa and the constant step dd. The nthn^{th} term identifies the value at a specific position, while the sum aggregates all values up to that position.

Problem 2:

In a GP, the 3rd3^{rd} term is 1212 and the 6th6^{th} term is 9696. Find the first term aa and the common ratio rr.

Solution:

  1. Express terms using the formula an=arn−1a_n = ar^{n-1}: ar2=12ar^2 = 12 (Equation 1) ar5=96ar^5 = 96 (Equation 2)
  2. Divide Equation 2 by Equation 1: ar5ar2=9612  ⟹  r3=8  ⟹  r=2\frac{ar^5}{ar^2} = \frac{96}{12} \implies r^3 = 8 \implies r = 2.
  3. Substitute r=2r = 2 back into Equation 1: a(2)2=12  ⟹  4a=12  ⟹  a=3a(2)^2 = 12 \implies 4a = 12 \implies a = 3.

Explanation:

For GP problems involving unknown aa and rr, set up a system of equations by dividing the higher term by the lower term to isolate the common ratio rr.