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Number and Algebra - Sequences and series

Grade 11IB_AI

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. This constant is called the common difference, dd.

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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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The sum of the terms in a sequence is called a series. Arithmetic and geometric series have specific formulae for the sum of the first nn terms (SnS_n).

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Sigma notation (∑\sum) is used to represent the sum of a sequence. For example, ∑i=1nui\sum_{i=1}^{n} u_i represents u1+u2+...+unu_1 + u_2 + ... + u_n.

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In financial applications, simple interest follows an arithmetic pattern, whereas compound interest follows a geometric pattern where the future value is FV=PV(1+r)nFV = PV(1 + r)^n.

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Linear depreciation is modeled by an arithmetic sequence with a negative common difference dd, while reducing balance depreciation is modeled by a geometric sequence with 0<r<10 < r < 1.

📐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=u1rn−1u_n = u_1 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}

FV=PV×(1+r100k)knFV = PV \times (1 + \frac{r}{100k})^{kn}

💡Examples

Problem 1:

Find the 15th15^{th} term and the sum of the first 15 terms of the arithmetic sequence: 7,11,15,...7, 11, 15, ...

Solution:

Identify u1=7u_1 = 7 and d=11−7=4d = 11 - 7 = 4. For the 15th15^{th} term: u15=7+(15−1)×4=7+56=63u_{15} = 7 + (15 - 1) \times 4 = 7 + 56 = 63. For the sum: S15=152(7+63)=7.5×70=525S_{15} = \frac{15}{2}(7 + 63) = 7.5 \times 70 = 525.

Explanation:

We use the nthn^{th} 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 total.

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 10 terms.

Solution:

Using un=u1rn−1u_n = u_1 r^{n-1}: 24=3×r4−1⇒24=3r3⇒r3=8⇒r=224 = 3 \times r^{4-1} \Rightarrow 24 = 3r^3 \Rightarrow r^3 = 8 \Rightarrow r = 2. Now 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 \times 1023 = 3069.

Explanation:

First, solve for rr by substituting the known terms into the general formula for a geometric sequence, then use the sum formula for a geometric series.

Problem 3:

An investment of P = 5000 is made at an interest rate of 4%4\% per annum, compounded quarterly. Calculate the value of the investment after 3 years.

Solution:

We use the compound interest formula where PV=5000PV = 5000, r=4r = 4, k=4k = 4 (quarterly), and n=3n = 3. FV=5000×(1+4100×4)4×3FV = 5000 \times (1 + \frac{4}{100 \times 4})^{4 \times 3} FV=5000×(1+0.01)12=5000×(1.01)12≈5634.13FV = 5000 \times (1 + 0.01)^{12} = 5000 \times (1.01)^{12} \approx 5634.13. The value is P 5634.13.

Explanation:

The formula FV=PV(1+r100k)knFV = PV(1 + \frac{r}{100k})^{kn} accounts for the compounding periods per year (kk). Quarterly compounding means k=4k=4.