Geometric Sequences, Series and Limiting Sums

A geometric sequence (GP) is one where each term is obtained by multiplying the previous term by a fixed constant $r$, the common ratio. Formally: $r = \dfrac{T_n}{T_{n-1}}$ is the same for every consecutive pair.

Testing whether a sequence is geometric: check that all successive ratios are equal. If $T_2/T_1 = T_3/T_2 = T_4/T_3 = \cdots = r$, the sequence is a GP.

Behaviour by ratio value:

• $r > 1$: terms grow without bound (exponential growth)
• $0 < r < 1$: terms decrease toward 0 (exponential decay)
• $r < 0$: terms alternate in sign, magnitude may grow or decay
• $r = 1$: all terms equal (constant sequence)
• $r = -1$: terms alternate between $a$ and $-a$

Contrast with AP: an AP adds a constant; a GP multiplies by a constant. This means $T_n$ in an AP is a linear function of $n$, while $T_n$ in a GP is an exponential function of $n$.

GP: $r = T_n / T_{n-1}$ is constant for all $n \geq 2$.; Test: compute $T_2/T_1$, $T_3/T_2$, equal means GP.

Pause, copy the GP definition: common ratio $r = T_n / T_{n-1}$ is constant; test by computing $T_2/T_1$ and $T_3/T_2$ (must be equal); $r > 1$ grows, $0 < r < 1$ decays, $r < 0$ alternates, into your book.

We just saw that a GP has a constant ratio $r = T_n / T_{n-1}$. That raises a question: instead of multiplying through term by term, can we jump directly to any position, and given two terms $T_m$ and $T_n$, can we find both $a$ and $r$ without listing all terms? This card answers it → $T_n = ar^{n-1}$; to find $a$ and $r$ from two known terms, divide to eliminate $a$ and solve $r^{n-m} = T_n/T_m$.

Starting from $T_1 = a$ and multiplying by $r$ each time: $T_1 = a$, $T_2 = ar$, $T_3 = ar^2$..., and by induction $T_n = ar^{n-1}$. Because $r^{n-1}$ is an exponential function of $n$, so is $T_n$.

Worked example 1, finding a term and solving for $n$:

GP: $3,\ 6,\ 12,\ \ldots$   $a = 3$, $r = 2$.

• Find $T_8$: $T_8 = 3 \cdot 2^7 = 3 \cdot 128 = 384$.
• Which term equals 384? Set $3 \cdot 2^{n-1} = 384 \Rightarrow 2^{n-1} = 128 = 2^7 \Rightarrow n-1 = 7 \Rightarrow n = 8$. ✓
• Which is the first term exceeding 1000? $3 \cdot 2^{n-1} > 1000 \Rightarrow 2^{n-1} > 333.3 \Rightarrow n-1 > \log_2(333.3) \approx 8.38 \Rightarrow n \geq 10$ (so $T_{10} = 3 \cdot 2^9 = 1536$).

Worked example 2, finding $a$ and $r$ from two terms:

$T_3 = 12$ and $T_6 = 96$. Find $a$ and $r$.

• $T_3 = ar^2 = 12$ and $T_6 = ar^5 = 96$.
• Divide: $\dfrac{ar^5}{ar^2} = r^3 = \dfrac{96}{12} = 8 \Rightarrow r = 2$.
• Back-substitute: $a \cdot 4 = 12 \Rightarrow a = 3$.
• Check: $T_1 = 3,\ T_2 = 6,\ T_3 = 12,\ T_4 = 24,\ T_5 = 48,\ T_6 = 96$ ✓

$T_n = ar^{n-1}$ where $a = T_1$ and $r$ = common ratio.; To find $a$ and $r$ from $T_m$ and $T_n$: divide to get $r^{n-m} = T_n/T_m$, take root, back-substitute.

Pause, copy $T_n = ar^{n-1}$ and the method for finding $a$ and $r$ from two known terms: divide $T_n$ by $T_m$ to get $r^{n-m}$, take roots, back-substitute for $a$, into your book.

We just saw that $T_n = ar^{n-1}$ finds any individual term. That raises a question: to sum the first $n$ terms of a GP, is there a shortcut, and how do you derive it rather than just memorise it? This card answers it → write $S_n$, multiply by $r$, subtract to cancel all interior terms, then solve: $S_n = \dfrac{a(1-r^n)}{1-r}$.

The GP sum formula is derived by a clever algebraic trick: write $S_n$, multiply by $r$, then subtract. Almost everything cancels.

Derivation (NESA requires you to know this):

• $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$   ...(1)
• $rS_n = \phantom{a + {}}ar + ar^2 + \cdots + ar^{n-1} + ar^n$   ...(2)
• Subtract (1) from (2): $rS_n - S_n = ar^n - a$
• $S_n(r-1) = a(r^n - 1)$, so $S_n = \dfrac{a(r^n-1)}{r-1}$ for $r \neq 1$.
• Equivalently: $S_n = \dfrac{a(1-r^n)}{1-r}$ (multiply top and bottom by $-1$).

Which form to use: both are equivalent, but:

• $\dfrac{a(1-r^n)}{1-r}$: numerator $(1-r^n) > 0$ when $r < 1$, convenient for $r < 1$.
• $\dfrac{a(r^n-1)}{r-1}$: numerator $(r^n-1) > 0$ when $r > 1$, convenient for $r > 1$.

Special case $r = 1$: all $n$ terms equal $a$, so $S_n = na$.

$S_n = \dfrac{a(1-r^n)}{1-r} = \dfrac{a(r^n-1)}{r-1}$ for $r \neq 1$. Use whichever keeps the numerator positive.; Special case $r = 1$: $S_n = na$.

Pause, copy the GP sum formula $S_n = \dfrac{a(1-r^n)}{1-r}$ (use when $r < 1$) or $\dfrac{a(r^n-1)}{r-1}$ (use when $r > 1$) and the special case $r = 1$: $S_n = na$, into your book.

We just saw the GP sum $S_n = \dfrac{a(1-r^n)}{1-r}$ for finite $n$. That raises a question: when $|r| < 1$, each term is smaller than the last, does the sum grow forever, or does it settle at a finite value as $n \to \infty$? This card answers it → since $r^n \to 0$ when $|r| < 1$, the formula gives $S_\infty = \dfrac{a}{1-r}$; when $|r| \geq 1$, no finite sum exists.

When $|r| < 1$, the terms of a GP shrink toward zero. As $n \to \infty$, $r^n \to 0$, so the sum formula simplifies to a finite value: $S_\infty = \dfrac{a}{1-r}$. When $|r| \geq 1$, no such limit exists.

Derivation of the limiting sum:

• Start with $S_n = \dfrac{a(1-r^n)}{1-r}$.
• If $|r| < 1$: as $n \to \infty$, $r^n \to 0$.
• So $S_n \to \dfrac{a(1-0)}{1-r} = \dfrac{a}{1-r}$.

Classic example: $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$: $a = 1$, $r = 0.5$.
$S_\infty = \dfrac{1}{1-0.5} = \dfrac{1}{0.5} = 2$ ✓

Repeating decimals as fractions: $0.\overline{3} = 0.3 + 0.03 + 0.003 + \cdots$: $a = 0.3$, $r = 0.1$.
$S_\infty = \dfrac{0.3}{1-0.1} = \dfrac{0.3}{0.9} = \dfrac{1}{3}$ ✓

Finding $r$ from $a$ and $S_\infty$:
If $a = 6$ and $S_\infty = 9$: $\dfrac{6}{1-r} = 9 \Rightarrow 1 - r = \dfrac{6}{9} = \dfrac{2}{3} \Rightarrow r = \dfrac{1}{3}$.

$S_\infty = \dfrac{a}{1-r}$ exists only when $|r| < 1$ (i.e. $-1 < r < 1$).; Derivation: in $S_n = \dfrac{a(1-r^n)}{1-r}$, take $n \to \infty$ and use $r^n \to 0$.

Pause, copy the limiting sum $S_\infty = \dfrac{a}{1-r}$ (exists only when $|r| < 1$) and the derivation logic (set $r^n \to 0$ in the finite sum formula) into your book.

We just saw the limiting sum $S_\infty = \dfrac{a}{1-r}$ for $|r| < 1$. That raises a question: when a population grows by 3% annually or a drug dose halves every 6 hours, how do you set up the GP model correctly, especially with the "after $k$ periods" indexing trap? This card answers it → growth: $r = 1 + \text{rate}$; decay: $r = 1 - \text{rate}$; "after $k$ periods" means $T_{k+1}$ if $T_1$ is the starting value.

Many real-world quantities change by a fixed percentage each period, populations, investments, radioactive decay, drug concentration in the bloodstream. These are modelled by GPs because each value is a fixed multiple of the previous one.

Setting up the model, identify $a$ and $r$:

• $a$ = initial amount (the starting value, $T_1$)
• Growth at rate $p\%$ per period: $r = 1 + p/100$ (e.g. 5% growth $\Rightarrow r = 1.05$)
• Decay at rate $p\%$ per period: $r = 1 - p/100$ (e.g. 20% decay $\Rightarrow r = 0.8$)

Worked example, bacteria colony:

A colony starts at 500 bacteria and triples every hour. (a) Value after 6 hours. (b) When does it first exceed 1,000,000?

• $a = 500$, $r = 3$. $T_n = 500 \cdot 3^{n-1}$.
• (a) After 6 hours means $n = 7$ (hour 0 is start, hour 6 is $T_7$): $T_7 = 500 \cdot 3^6 = 500 \cdot 729 = 364{,}500$ bacteria.
• (b) $500 \cdot 3^{n-1} > 10^6 \Rightarrow 3^{n-1} > 2000 \Rightarrow (n-1)\log 3 > \log 2000$
  $n-1 > \dfrac{\log 2000}{\log 3} \approx \dfrac{3.301}{0.477} \approx 6.92 \Rightarrow n \geq 8$.
  So at $n = 8$ (after 7 hours): $T_8 = 500 \cdot 3^7 = 500 \cdot 2187 = 1{,}093{,}500 > 1{,}000{,}000$. ✓

Sum applications: GP sums arise when the context accumulates over time, total drug doses, total distance travelled by a bouncing ball, total depreciation.

Growth model: $a$ = initial, $r = 1 + \text{rate}$. Decay model: $r = 1 - \text{rate}$.; Apply $T_n = ar^{n-1}$ with careful indexing: "after $k$ periods" = $T_{k+1}$ if $T_1$ is the start.

Pause, copy the GP application setup: growth $r = 1 + \text{rate}$, decay $r = 1 - \text{rate}$; $T_n = ar^{n-1}$ with the indexing rule that "after $k$ periods" = $T_{k+1}$, into your book.