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Algebra - Sequences (Arithmetic and Geometric)

Grade 10IGCSE

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

🔑Concepts

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Definition of a sequence: A list of numbers following a specific mathematical pattern or rule.

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Arithmetic Sequence (Linear): A sequence where the difference between consecutive terms is constant, known as the common difference (d).

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

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Convergent Geometric Series: A geometric sequence where the sum approaches a specific value as n approaches infinity (only occurs when |r| < 1).

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Term Position (n): The position of a term in a sequence, which must always be a positive integer (1, 2, 3...).

📐Formulae

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

Sum of first n arithmetic terms: 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 L is the last term

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

Sum of first n geometric terms: Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} for r>1r > 1 or Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r} for r<1r < 1

Sum to infinity (Geometric): S∞=a1−rS_{\infty} = \frac{a}{1 - r} (valid only if ∣r∣<1|r| < 1)

💡Examples

Problem 1:

Find the 15th term of the arithmetic sequence: 7, 11, 15, 19, ...

Solution:

u15=63u_{15} = 63

Explanation:

Identify a=7a = 7 and d=11−7=4d = 11 - 7 = 4. Use the formula un=a+(n−1)du_n = a + (n-1)d. Substitute the values: u15=7+(15−1)4=7+(14×4)=7+56=63u_{15} = 7 + (15-1)4 = 7 + (14 \times 4) = 7 + 56 = 63.

Problem 2:

In a geometric sequence, the first term is 5 and the common ratio is 2. Find the sum of the first 6 terms.

Solution:

S6=315S_6 = 315

Explanation:

Identify a=5,r=2,n=6a = 5, r = 2, n = 6. Use the sum formula Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}. Substitute: S6=5(26−1)2−1=5(64−1)1=5×63=315S_6 = \frac{5(2^6 - 1)}{2 - 1} = \frac{5(64 - 1)}{1} = 5 \times 63 = 315.

Problem 3:

Calculate the sum to infinity of the geometric sequence: 10, 5, 2.5, 1.25, ...

Solution:

S∞=20S_{\infty} = 20

Explanation:

Identify a=10a = 10 and r=5/10=0.5r = 5/10 = 0.5. Since ∣0.5∣<1|0.5| < 1, the sum to infinity exists. Use S∞=a1−r=101−0.5=100.5=20S_{\infty} = \frac{a}{1 - r} = \frac{10}{1 - 0.5} = \frac{10}{0.5} = 20.