Mini-Lesson

Number & Algebra

This mini-lesson covers the SL Number & Algebra toolkit for AI: approximation & percentage error, standard form, arithmetic & geometric sequences and series, financial maths (compound interest, depreciation, annuities and loan amortisation) and logarithms.

AI flavour: Number & Algebra in Applications is about using these tools on real money and measurement problems with your GDC — not abstract proof.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Every number on the calculation screens has been re-derived and checked. Press Start when you are ready.

Approximation & error

Rounding, bounds & percentage error

Every measurement is approximate. If v_A is an approximate (measured) value and v_E is the exact value, the percentage error is:

ε = |v_A − v_E| ÷ |v_E| × 100%always take the exact value on the bottom

Worked example

A ruler reads 2.5 cm; the true length is 2.4 cm.

ε = |2.5 − 2.4| ÷ 2.4 × 100 = 0.1 ÷ 2.4 × 100 = 4.17% (3 s.f.)

Watch out: it is |approx − exact| over the exact value, not over the approximate one.

Calculate

Your turn — percentage error

1A stopwatch records a race as 12.5 s; the electronic timing gives the exact time as 12.4 s. Find the percentage error to 3 significant figures.

%

Hint: |12.5 − 12.4| ÷ 12.4 × 100.

Standard form

Standard form (scientific notation)

Standard form writes a number as a × 10^k where 1 ≤ a < 10 and k is an integer. It is essential for very large and very small quantities.

384 400 km (Earth–Moon) = 3.844 × 10⁵ km
0.000 042 = 4.2 × 10⁻⁵
Multiply: (3 × 10⁸) × (2 × 10⁻³) = 6 × 10⁵ = 600 000

GDC tip: your calculator writes 6E5 or 6×10^5 — read the exponent carefully.

Quick check

Standard form

?Written in standard form a × 10^k, what is 384 400?

Calculate

Your turn — standard form product

2Evaluate (3 × 10⁸) × (2 × 10⁻³) and give the ordinary (non-standard-form) number.

Hint: 3 × 2 = 6 and 10⁸ × 10⁻³ = 10⁵.

Sequences & series

Arithmetic sequences & series

An arithmetic sequence adds a fixed common difference d each term.

u_n = u₁ + (n−1)dS_n = n⁄2 (2u₁ + (n−1)d)

Worked example

u₁ = 5, d = 3.

u₁₀ = 5 + 9×3 = 32

S₁₀ = 10⁄2 (2×5 + 9×3) = 5 × 37 = 185

Calculate

Your turn — arithmetic term

3For the arithmetic sequence with u₁ = 5 and common difference d = 3, find the 10th term u₁₀.

Hint: u₁₀ = 5 + (10−1)×3.

Sequences & series

Geometric sequences & series

A geometric sequence multiplies by a fixed common ratio r each term.

u_n = u₁ rⁿ⁻¹S_n = u₁(rⁿ − 1) ÷ (r − 1)

Worked example

u₁ = 3, r = 2.

u₈ = 3 × 2⁷ = 3 × 128 = 384

S₈ = 3(2⁸ − 1) ÷ (2 − 1) = 3 × 255 = 765

Calculate

Your turn — geometric sum

4For the geometric sequence with u₁ = 3 and r = 2, find the sum of the first 8 terms S₈.

Hint: S₈ = 3(2⁸ − 1) ÷ (2 − 1).

Sort it

Which kind of growth?

Tap a description, then the type of sequence or tool it belongs to.

➕ Arithmetic

✖️ Geometric

💷 Financial

Financial maths

Compound interest & depreciation

Money grown at compound interest uses:

A = P(1 + r⁄n)^ntP = principal · r = annual rate · n = compounds per year · t = years

Worked example — monthly compounding

P = $2000, r = 0.03 (3%), n = 12, t = 5.

A = 2000(1 + 0.03⁄12)⁶⁰ = 2000 × 1.0025⁶⁰ = $2323.23

Depreciation uses a negative rate: value 20 000 losing 15%/yr for 3 yr = 20000 × 0.85³ = 12 282.50.

Calculate

Your turn — compound interest

5$2000 is invested at 3% per year compounded monthly (n = 12) for 5 years. Use A = P(1 + r⁄n)^nt to find the final amount, to the nearest cent.

$

Hint: 2000 × (1 + 0.03⁄12)^12×5.

Financial maths

Annuities & loan amortisation

A repeated equal payment PMT (an annuity) has future value, and a loan is amortised by equal repayments. With periodic rate i and N periods:

FV = PMT · ((1+i)^N − 1) ÷ iPMT_loan = L·i ÷ (1 − (1+i)^−N)

Real example: a $15 000 loan at 6%/yr compounded monthly over 5 years (i = 0.005, N = 60) needs a monthly repayment of $289.99. Saving $200/month at 5%/yr monthly for 10 years grows to $31 056.46. Use your GDC finance solver.

Logarithms

A logarithm answers “what power?”: log_bx = y means b^y = x. AI uses logs mainly to solve exponential (growth/decay) equations.

log₂32 = 5 because 2⁵ = 32laws: log(ab)=log a+log b · log(a⁄b)=log a−log b · log(aⁿ)=n log a

Quick check

Financial vocabulary

?A borrower repays a fixed monthly amount that gradually clears both interest and the original loan. This process is called:

Quick check

Spot the sequence

?The sequence 4, 12, 36, 108, … is best described as:

Quick check

Percentage error idea

?In the percentage-error formula, which value goes on the bottom (the denominator)?

Match it

Match the formula to its use

Tap an item on the left, then its match on the right.

Item

Match

Recap

The big ideas to take away

Percentage error: |approx − exact| ÷ |exact| × 100%

Standard form: a × 10^k with 1 ≤ a < 10

Arithmetic: u_n = u_1 + (n−1)d ; S_n = n⁄2(2u_1 + (n−1)d)

Geometric: u_n = u_1 r^(n−1) ; S_n = u_1(r^n − 1)⁄(r − 1)

Compound interest: A = P(1 + r⁄n)^(nt) ; depreciation uses a negative rate

Finance: annuities/amortisation via GDC finance solver; logs solve exponential equations

You have worked through the whole topic. Press Finish to see your score.

🏆

Mini-lesson complete!

⭐⭐⭐

You have covered the SL Number & Algebra toolkit for AI. 🎉

Your stars: 0 / 0

Next: test yourself in the Evaluate stage, then lock it in with Verify.

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