Real growth, decay and repeating processes never arrive labelled "arithmetic" or "geometric" — the real skill is reading a word problem and recognising which structure is hiding inside it.
10 min read4 worked examples
Choosing the right model
Every technique from this topic so far — arithmetic sequences and series, geometric sequences and series, and recurrence relations — is really a different answer to the same question: how does one term relate to the next? A modelling question rarely says which technique to use; it describes a real situation, and the first job is to translate that description into $a$, $d$ or $r$ (or a full recurrence), before any formula gets touched.
Spotting the structure
A constant amount added or subtracted every period (e.g. "£1,200 less each year") signals an arithmetic model. A constant percentage or fixed multiplying factor every period (e.g. "grows by 8% each hour") signals a geometric model. A fixed amount combined with a percentage every period — a regular deposit that also earns interest, or a drug dose that partially clears before the next one — doesn't fit either pattern alone, and is best modelled with a recurrence relation.
Sequences versus series in modelling
It's also worth pausing on what's actually being asked for. A question about the value of one term — "what's the balance after 5 years?" — is a sequence question, answered with $u_n$. A question about a running total across many periods — "what's been paid in altogether?", "how many bricks in the whole pile?" — is a series question, answered with $S_n$. Regular-saving problems are a common trap here: each individual deposit grows geometrically on its own, but the account balance is the sum of every deposit's growth, which makes it a series, not a single sequence term.
Worked example 1
An arithmetic model: straight-line depreciation
A company buys a machine for £18,000. Its value decreases by £1,200 each year, forming an arithmetic sequence. Find the value of the machine after 5 years, and determine after how many complete years the value first falls below £6,000.
MethodIdentify a and d from the context, then apply the nth-term formula
1
Let $u_n$ be the value after $n$ years. The value after 1 year is $18000-1200=16800$, so $a=16800$ and $d=-1200$.
Translate "decreases by £1,200 each year" into a common difference.
2
$u_5 = a+4d = 16800+4(-1200) = 12000$, so the machine is worth £12,000 after 5 years.
Check the boundary: $u_{10}=16800-1200(9)=6000$ exactly (not below), while $u_{11}=16800-1200(10)=4800$ (below).
$n>10$ means $n=10$ itself doesn't satisfy the strict inequality — always verify the integer boundary directly.
5
The value first falls below £6,000 after 11 complete years.
Final answer.
▪ solved
Worked example 2
A geometric model: exponential growth and logarithms
A population of bacteria is modelled by $u_n=ar^{n-1}$, where $u_1=500$ is the population at the start and $r=1.08$ reflects 8% hourly growth. Find the least number of complete hours after which the population first exceeds 2000.
MethodSet up the inequality, take logs to bring n down, then convert back to elapsed hours
Divide both sides by 500 to isolate the power of $r$.
2
Take logs of both sides: $(n-1)\ln(1.08)>\ln4$.
Logarithms bring the exponent $n-1$ down to a multiplier, since $\ln(1.08)>0$ the inequality direction is unchanged.
3
$n-1 > \dfrac{\ln4}{\ln1.08} \approx 18.01 \Rightarrow n>19.01$, so the least integer is $n=20$.
Divide, then round up to the next whole term number since $n$ must be a positive integer.
4
Since $u_1$ is the population at the start (0 hours elapsed), $u_n$ corresponds to $n-1$ hours elapsed. Checking: $u_{19}\approx1998.0<2000$ (18 hours), while $u_{20}\approx2157.9>2000$ (19 hours).
Verify the boundary directly, and convert the term number back into hours elapsed.
5
The population first exceeds 2000 after 19 complete hours.
Final answer, stated in the units the question asked for.
▪ solved
Term number versus elapsed time
Whenever $u_1$ represents a starting value at "time zero," the term number $n$ and the elapsed time differ by exactly one — a frequent source of off-by-one errors in modelling questions. Always state explicitly what $u_1$ represents before converting a term number back into a real-world answer.
Worked example 3
A series model: regular savings with compound interest
At the start of each year, Sarah pays £1000 into a savings account paying 5% annual interest, compounded once a year on the balance at the end of each year. Find the total value of her savings at the end of the 10th year, to the nearest penny.
MethodModel each deposit's own growth, then sum them as a geometric series
1
Her 1st deposit sits in the account for the full 10 years, becoming $1000(1.05)^{10}$. Her 2nd deposit grows for 9 years, becoming $1000(1.05)^9$. This continues down to her 10th deposit, which grows for just 1 year, becoming $1000(1.05)^1$.
Each deposit is a separate geometric sequence, since each earns compound interest independently from the moment it's paid in.
2
The total balance is the sum $1000(1.05)+1000(1.05)^2+\cdots+1000(1.05)^{10}$ — a geometric series with $a=1000(1.05)=1050$, $r=1.05$, and $n=10$ terms.
The running total across all 10 deposits is exactly a geometric series sum.
Each individual deposit follows a geometric sequence on its own, but the question asks for the total balance — the sum of ten separate, differently-aged deposits. Recognising that a running total across many periods calls for the sum formula, not the $n$th-term formula, is the key modelling step here.
Worked example 4
A recurrence model: the long-run steady state · Challenge
A patient takes a 100 mg dose of a drug at the start of each day. Overnight, 30% of the drug in the body is eliminated before the next dose. If $u_n$ is the amount of drug (in mg) in the body immediately after the $n$th dose, then $u_{n+1}=0.7u_n+100$. Show that, in the long run, $u_n$ approaches a fixed limit $L$, and find its exact value.
MethodAssume a limit exists, substitute it into the recurrence, then solve for it
1
If $u_n$ settles on a limiting value $L$ as $n\to\infty$, then eventually $u_{n+1}\approx u_n\approx L$, so $L$ must satisfy the recurrence with both sides replaced by $L$: $L=0.7L+100$.
A fixed point of the recurrence is a value that, once reached, never changes.
Since the multiplier $0.7$ satisfies $|0.7|<1$, each step shrinks the gap between $u_n$ and $L$ by a factor of $0.7$ — the same shrinking mechanism that makes a geometric series with $|r|<1$ converge to a sum to infinity.
This is what guarantees a limit genuinely exists, rather than the sequence growing or oscillating forever.
4
So in the long run, the amount of drug in the body approaches $L=\dfrac{1000}{3}\approx333.3$ mg.
Final answer, as an exact fraction and a rounded decimal.
▪ solved
The fixed-point method
Substituting $u_{n+1}=u_n=L$ into a recurrence relation and solving for $L$ finds any limiting value directly, without generating term after term by hand. It only works when the recurrence is actually converging — here, that's guaranteed by the multiplier $0.7$ having magnitude less than 1, exactly as $|r|<1$ guarantees convergence for a geometric series.
Common mistake
Forcing a context into the wrong family of model — treating "a regular deposit that also earns interest" as purely geometric (which ignores the repeated deposits) or purely arithmetic (which ignores the compounding), when it actually needs a recurrence relation or a sum of a series to capture both effects at once. Always check whether a scenario has one changing quantity (arithmetic or geometric) or two combined effects (recurrence), and whether the question wants a single value ($u_n$) or a running total ($S_n$).
Check yourself
At the start of each month, Tom deposits £200 into a savings account paying 0.4% interest per month, compounded monthly on the balance at the end of each month. Find the total value of his savings at the end of the 24th month, to the nearest penny. (Hint: each deposit grows for a different number of months — model the total as a geometric series.)
Method
1
The 1st deposit grows for 24 months, becoming $200(1.004)^{24}$; the 24th deposit grows for just 1 month, becoming $200(1.004)^1$.
Each of the 24 deposits earns compound interest for a different number of months.
2
The total is a geometric series with $a=200(1.004)=200.8$, $r=1.004$, $n=24$ terms: $S_{24}=\dfrac{a(r^{24}-1)}{r-1}$.
Set up the sum formula from the identified values.