Geometric Progression — chapter notes

• A geometric progression (GP) is a list in which each term is the term before it multiplied by the same non-zero number $r$, the common ratio.
• $r = \dfrac{T_{n+1}}{T_n}$: divide any term by the term before it. A list is a GP only if every such ratio is the same.
• $r$ can be a fraction (terms shrink) or negative (signs alternate). No term of a GP is $0$; if $r = 1$ every term is equal.

Example: $2, -6, 18, -54$: ratios $-3, -3, -3$, so it is a GP with $r = -3$; the next term is $162$.

The terms are $a, ar, ar^{2}, ar^{3}, \ldots$, so the $n$th term is $T_n = ar^{n-1}$: the power of $r$ is one less than the term number.

• To find a term: substitute $a$, $r$ and $n$.
• To find which term equals a value: solve $ar^{n-1} =$ value by writing both sides as powers of $r$.

Examples: the $8$th term of $81, -27, 9, \ldots$ is $81 \times \left(-\dfrac{1}{3}\right)^{7} = -\dfrac{1}{27}$; in $6, 18, 54, \ldots$, $6 \times 3^{n-1} = 4374$ gives $3^{n-1} = 729 = 3^{6}$, so $n = 7$.

1. Write the given terms as $ar^{p-1}$ and $ar^{q-1}$.
2. Divide: $a$ cancels and a single power of $r$ remains.
3. Take the root. An even power has two real roots, $\pm$; check both. An odd power has one real root, with the sign of the number.
4. Find $a$ from either term, and check by listing a few terms.

Examples: $T_3 = 36$, $T_6 = 972$: $r^{3} = 27$, $r = 3$, $a = 4$. $T_2 = 6$, $T_4 = 54$: $r^{2} = 9$, so $r = 3$ ($a = 2$) or $r = -3$ ($a = -2$).

• If $a$, $b$, $c$ are consecutive terms of a GP, $\dfrac{b}{a} = \dfrac{c}{b}$, so $b^{2} = ac$. For positive terms, $b = \sqrt{ac}$.
• Three unknown terms in GP with a known product: write them as $\dfrac{a}{r}$, $a$, $ar$; their product is $a^{3}$.
• From the end: start at the last term $l$ and divide by $r$ at each step: the $k$th term from the end is $l\left(\dfrac{1}{r}\right)^{k-1}$.

Examples: $x - 2$, $x + 1$, $x + 7$ in GP: $(x + 1)^{2} = (x - 2)(x + 7)$ gives $x = 5$ (terms $3, 6, 12$). In $5, 15, 45, \ldots, 10935$ the $3$rd term from the end is $10935 \times \left(\dfrac{1}{3}\right)^{2} = 1215$.

Write $S_n = a + ar + \cdots + ar^{n-1}$, multiply by $r$ and subtract: $rS_n - S_n = ar^{n} - a$.

• $r > 1$: $S_n = \dfrac{a\left(r^{n} - 1\right)}{r - 1}$.
• $r < 1$ (including negative $r$): $S_n = \dfrac{a\left(1 - r^{n}\right)}{1 - r}$ (the same value, written to keep top and bottom positive).
• $r = 1$: every term is $a$, so $S_n = na$.

Examples: $4 + 12 + 36 + \cdots$ ($6$ terms) $= 1456$; $96 + 48 + 24 + \cdots$ ($6$ terms) $= 189$; $2 - 4 + 8 - \cdots$ ($7$ terms) $= 86$.

• How many terms give a total: set $S_n$ equal to the total, simplify to $r^{n} =$ a number, and compare powers.
• A term from sums: $T_n = S_n - S_{n-1}$.
• Least number of terms to pass a value: work out consecutive sums and compare.

Examples: $7 + 21 + 63 + \cdots = 2548$ gives $3^{n} = 729$, so $n = 6$. If $S_n = 3\left(2^{n} - 1\right)$, then $T_5 = S_5 - S_4 = 93 - 45 = 48$.

• Growth by $p\%$ of the current value each year: $r = 1 + \dfrac{p}{100}$ (population, savings, a growing plant).
• Depreciation by $p\%$ of the current value each year: $r = 1 - \dfrac{p}{100}$ (a machine, a vehicle).
• Starting value $a$: the value after $k$ years is $ar^{k}$, the $(k + 1)$th term of the GP.
• Doubling, tripling or halving each step: $r = 2$, $3$ or $\dfrac{1}{2}$; a total over several steps is a sum $S_n$.

Examples: $25000$ growing at $4\%$: after $2$ years $25000 \times 1.04^{2} = 27040$. ₹1,60,000 falling $10\%$ a year: after $3$ years $160000 \times 0.9^{3} = 116640$ rupees.