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Algebra - Sequences and Series

Grade 12IGCSE

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 consecutive terms is constant (common difference, d).

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Geometric Progression (GP): A sequence where each term is found by multiplying the previous term by a constant (common ratio, r).

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Sigma Notation (Σ): A concise way to write the sum of a sequence of numbers.

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Convergent Series: A geometric series where the common ratio |r| < 1, allowing the sum to approach a finite limit as n approaches infinity.

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Divergent Series: A sequence that does not approach a finite limit.

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The nth term (general term): Represented as u_n or a_n, used to find any specific term in a sequence.

📐Formulae

un=a+(n−1)du_n = a + (n - 1)d (Arithmetic nth term)

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n - 1)d] (Sum of n terms in AP)

un=arn−1u_n = ar^{n-1} (Geometric nth term)

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

S∞=a1−r,∣r∣<1S_\infty = \frac{a}{1 - r}, |r| < 1 (Sum to infinity for convergent GP)

∑r=1nur=u1+u2+...+un\sum_{r=1}^{n} u_r = u_1 + u_2 + ... + u_n (Sigma Notation)

💡Examples

Problem 1:

Find the 15th term and the sum of the first 15 terms of the arithmetic sequence: 5, 9, 13, ...

Solution:

u15=61u_{15} = 61, S15=495S_{15} = 495

Explanation:

Here, the first term a=5a = 5 and the common difference d=9−5=4d = 9 - 5 = 4. Using the nth term formula: u15=5+(15−1)4=5+56=61u_{15} = 5 + (15-1)4 = 5 + 56 = 61. Using the sum formula: S15=152[2(5)+(15−1)4]=7.5[10+56]=7.5×66=495S_{15} = \frac{15}{2}[2(5) + (15-1)4] = 7.5[10 + 56] = 7.5 \times 66 = 495.

Problem 2:

A geometric progression has a first term of 18 and a second term of 6. Calculate the sum to infinity.

Solution:

S∞=27S_\infty = 27

Explanation:

First, find the common ratio r=u2u1=618=13r = \frac{u_2}{u_1} = \frac{6}{18} = \frac{1}{3}. Since ∣1/3∣<1|1/3| < 1, the series converges. Use the formula S∞=a1−rS_\infty = \frac{a}{1-r}. Substituting the values: S∞=181−1/3=182/3=18×32=27S_\infty = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 18 \times \frac{3}{2} = 27.

Problem 3:

Evaluate ∑r=15(2r+3)\sum_{r=1}^{5} (2^r + 3).

Solution:

77

Explanation:

Expand the sigma notation by substituting r=1,2,3,4,5r = 1, 2, 3, 4, 5: (21+3)+(22+3)+(23+3)+(24+3)+(25+3)=5+7+11+19+35=77(2^1+3) + (2^2+3) + (2^3+3) + (2^4+3) + (2^5+3) = 5 + 7 + 11 + 19 + 35 = 77. Alternatively, split into a GP sum and a constant sum: ∑2r+∑3=2(25−1)2−1+(5×3)=62+15=77\sum 2^r + \sum 3 = \frac{2(2^5-1)}{2-1} + (5 \times 3) = 62 + 15 = 77.