IB Maths AI HL Sequences & Series Paper 1 & 2 interest & growth ~7 min read

Applications of Sequences & Series

Worded problems rarely say “arithmetic” or “geometric” — you have to spot which model fits. A fixed amount added each time is arithmetic; a fixed factor multiplied each time is geometric. Once you have chosen, the standard formulae do the rest.

📘 What you need to know

Recognising an arithmetic model

Use an arithmetic model whenever a quantity changes by the same amount added or subtracted each step. The starting value is u1 and the fixed change is the common difference d.

Typical contexts include simple interest, cups stacked in a tapering pile, trees planted in fixed batches, and dots counted in a growing pattern.

Which model? Look at how the quantity changes ARITHMETIC add / subtract a fixed amount 100 → 110 → 120 → 130 +10   +10   +10 • simple interest • stacking cups / tins • planting in fixed batches • counting pattern dots GEOMETRIC multiply by a fixed ratio 100 → 110 → 121 → 133.1 ×1.1   ×1.1   ×1.1 • compound interest • population growth • bacterial growth • radioactive decay fixed amount → arithmetic  •  fixed factor → geometric
The test is the type of change: a constant amount added points to arithmetic; a constant factor multiplied points to geometric.
Arithmetic model — fixed amount added un = u1 + (n − 1)d Sn = n2(2u1 + (n − 1)d) use a term for one value, a sum for a cumulative total

Recognising a geometric model

Use a geometric model whenever a quantity is repeatedly multiplied or divided by the same factor. The starting value is u1 and the fixed factor is the common ratio r.

Typical contexts include compound interest, population and bacterial growth, and radioactive decay.

Geometric model — fixed factor multiplied un = u1rn−1 Sn = u1(rn − 1)r − 1 for an x% change: r = 1 + x/100 (growth) or 1 − x/100 (decay)

Interest, growth and decay

Simple interest adds the same amount each year — an arithmetic model. Compound interest multiplies by the same factor each year — a geometric model. Growth and decay work the same way: a percentage change per period gives a common ratio.

Watch the percentage: a 6% rise means multiply by 1.06; a 6% fall means multiply by 0.94. Convert the percentage into a ratio before using any formula.

🧭 Recipe — tackling an applications problem

  1. Identify the changing quantity and write down its starting value.
  2. Decide the model — a fixed amount added is arithmetic; a fixed factor multiplied is geometric.
  3. State u1 and d or r — for a percentage, turn it into a ratio first.
  4. Choose term or sumun for a single value, Sn for a cumulative total.
  5. Answer in context — add units, round sensibly and check the result is realistic.

Worked examples

WE 1

Arithmetic: a stacking pattern

Tins are stacked in rows. The bottom row has 28 tins, and each row above has 3 fewer tins. How many tins are in the 8th row?

fixed amount removed each row ⇒ arithmetic u1 = 28,   d = −3 find the 8th term u8 = 28 + (8 − 1)(−3) = 28 − 21 7 tins a decreasing pattern just means a negative common difference.
WE 2

Arithmetic: reaching a total

Maria saves money each week. In week 1 she saves $8, and each week she saves $3 more than the week before. In which week do her total savings first reach at least $200?

fixed amount added ⇒ arithmetic; u1 = 8, d = 3 a total ⇒ use Sn; set Sn = 200 n2(2(8) + (n − 1)(3)) = 200 3n2 + 13n − 400 = 0 ⇒ n ≈ 9.58 check whole weeks either side S9 = 180 (< 200),   S10 = 215 (≥ 200) week 10 a non-integer n means you test the whole numbers on each side.
WE 3

Simple interest

$2000 is invested at 5% simple interest per year — the interest each year is 5% of the original $2000. (a) Find the interest earned each year. (b) Find the total value of the investment after 6 years.

(a) simple interest: the same amount each year interest = 5% of 2000 = 0.05 × 2000 = $100 (b) $100 added each year ⇒ arithmetic value = 2000 + 6 × 100 = 2000 + 600 (a) $100 per year  ·  (b) $2600 simple interest adds a constant — the value grows arithmetically.
WE 4

Compound interest

$3000 is invested at 4% compound interest per year. Find the value of the investment after 5 years, to the nearest dollar.

compound interest: multiply by a fixed factor ⇒ geometric 4% growth ⇒ r = 1 + 0.04 = 1.04 value after 5 years 3000 × 1.045 = 3000 × 1.2167 = 3649.96… ≈ $3650 compound interest multiplies each year, so it grows geometrically.
WE 5

Geometric: a spreading total

A post is shared by 5 people on day 1. Each day, the number of people who newly share it is triple the previous day’s. (a) How many people newly share it on day 6? (b) Find the total number of shares over the first 6 days.

tripled each day ⇒ geometric; u1 = 5, r = 3 (a) a single day ⇒ use the term formula u6 = 5 × 35 = 5 × 243 = 1215 (b) a total ⇒ use the sum formula S6 = 5(36 − 1)3 − 1 = 5(728)2 (a) 1215  ·  (b) 1820 “on day 6” wants a term; “total over 6 days” wants a sum.
WE 6

Full question: radioactive decay

A radioactive sample has a mass of 80 mg and loses 15% of its mass each year. (a) Explain why the masses form a geometric sequence and state the common ratio. (b) Find the mass after 4 years. (c) Find the mass after 10 years, to 1 decimal place.

(a) each year the mass keeps 85% of its value multiplied by the same factor ⇒ geometric, r = 0.85 (b) mass after n years = 80 × 0.85n 80 × 0.854 = 80 × 0.522 = 41.76 (c) after 10 years 80 × 0.8510 = 15.749… (a) r = 0.85 · (b) 41.8 mg · (c) 15.7 mg a 15% loss means keeping 85%, so the decay ratio is 0.85, not 0.15.

💡 Top tips

âš  Common mistakes

That completes Sequences & Series. You now have the full toolkit — the language of terms and sums, sigma notation, arithmetic and geometric models, and the real-world applications that tie them together. The deciding question is always the same: is the change a fixed amount, or a fixed factor?

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