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Algebra - Arithmetic and geometric sequences and series-extended

Grade 9IB

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

🔑Concepts

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An Arithmetic Sequence is a sequence where the difference between consecutive terms is constant, known as the common difference dd.

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The nthn^{th} term of an arithmetic sequence is denoted as unu_n, where u1u_1 is the first term.

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A Geometric Sequence is a sequence where each term is found by multiplying the previous term by a constant called the common ratio rr.

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A Series is the sum of the terms of a sequence. It can be finite (SnS_n) or infinite (S∞S_{\infty}).

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A geometric series converges (has a finite sum to infinity) only if the absolute value of the common ratio is less than 1, i.e., ∣r∣<1|r| < 1.

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Sigma notation ∑\sum is used to represent the sum of a sequence over a specific range of terms.

📐Formulae

un=u1+(n−1)du_n = u_1 + (n - 1)d

Sn=n2(u1+un)=n2(2u1+(n−1)d)S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}(2u_1 + (n - 1)d)

un=u1⋅rn−1u_n = u_1 \cdot r^{n-1}

Sn=u1(rn−1)r−1=u1(1−rn)1−rS_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}

S∞=u11−r for ∣r∣<1S_{\infty} = \frac{u_1}{1 - r} \text{ for } |r| < 1

💡Examples

Problem 1:

Find the 20th20^{th} term and the sum of the first 2020 terms of the arithmetic sequence: 5,12,19,26,…5, 12, 19, 26, \dots.

Solution:

  1. Identify parameters: u1=5u_1 = 5 and d=12−5=7d = 12 - 5 = 7.
  2. Find u20u_{20}: u20=5+(20−1)⋅7=5+133=138u_{20} = 5 + (20 - 1) \cdot 7 = 5 + 133 = 138
  3. Find S20S_{20}: S20=202(5+138)=10⋅143=1430S_{20} = \frac{20}{2}(5 + 138) = 10 \cdot 143 = 1430

Explanation:

We use the general term formula un=u1+(n−1)du_n = u_1 + (n-1)d to find the specific term and the sum formula Sn=n2(u1+un)S_n = \frac{n}{2}(u_1 + u_n) for the series.

Problem 2:

A geometric sequence has u1=3u_1 = 3 and u4=24u_4 = 24. Find the common ratio rr and the sum of the first 1010 terms.

Solution:

  1. Find rr: u4=u1⋅r4−1  ⟹  24=3⋅r3  ⟹  r3=8  ⟹  r=2u_4 = u_1 \cdot r^{4-1} \implies 24 = 3 \cdot r^3 \implies r^3 = 8 \implies r = 2
  2. Find S10S_{10}: S10=3(210−1)2−1=3(1024−1)=3(1023)=3069S_{10} = \frac{3(2^{10} - 1)}{2 - 1} = 3(1024 - 1) = 3(1023) = 3069

Explanation:

First, use the nthn^{th} term formula to solve for the unknown ratio rr. Then, apply the geometric series sum formula.

Problem 3:

Determine if the geometric series 10+5+2.5+…10 + 5 + 2.5 + \dots converges. If it does, find the sum to infinity.

Solution:

  1. Identify u1=10u_1 = 10 and r=510=0.5r = \frac{5}{10} = 0.5.
  2. Check convergence: Since ∣0.5∣<1|0.5| < 1, the series converges.
  3. Calculate S∞S_{\infty}: S∞=101−0.5=100.5=20S_{\infty} = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20

Explanation:

A geometric series converges if its ratio is between −1-1 and 11. The sum to infinity formula S∞=u11−rS_{\infty} = \frac{u_1}{1-r} provides the limit of the sum as nn approaches infinity.