Geometric Sequences and Means — Study Notes

Geometric Sequence

  • Definition: In a geometric sequence, every term is obtained by multiplying or dividing the preceding term by a fixed number, called the common ratio, denoted as $r$.
  • Notation:
    • $a_n$ = nth term of the sequence
    • $a_1$ = first term
    • $n$ = term number
    • $r$ = common ratio
  • General term:
    • a<em>n=a</em>1 r n−1a<em>n = a</em>1 \, r^{\,n-1}
  • How to identify a geometric sequence:
    • Check whether the ratio of consecutive terms is constant: $\dfrac{a{n}}{a{n-1}} = r$ (same for all $n$).
  • Key concepts from the transcript:
    • In a geometric sequence, each term is obtained by multiplying or dividing the previous term by a definite number (the common ratio).
    • The notations for terms used: $an$ (nth term), $a1$ (first term), $n$ (term position), $r$ (common ratio).
  • Example (finding $a_1$ and $r$ from given terms):
    • Given $a2 = 20$, $a6 = 320$ for a geometric sequence.
    • Since $a2 = a1 r$ and $a6 = a1 r^5$, divide: $\dfrac{a6}{a2} = \dfrac{a1 r^5}{a1 r} = r^4 = \dfrac{320}{20} = 16$.
    • Therefore $r = 16^{1/4} = 2$.
    • Then $a1 = \dfrac{a2}{r} = \dfrac{20}{2} = 10$.
    • Answer: First term $a_1 = 10$, common ratio $r = 2$.
  • Relationship to arithmetic sequences:
    • Geometric sequences involve a constant multiplicative change (ratio $r$), whereas arithmetic sequences involve a constant additive change (difference $d$).
  • Important related concept: Determine whether a sequence is geometric OR arithmetic by checking if consecutive ratios are constant (GP) or consecutive differences are constant (AP).
  • Common tasks (from exercises in the transcript):
    • Determine if a sequence is geometric; if yes, find the common ratio.
    • Identify whether a sequence is geometric or arithmetic.
  • Notation recap: For a geometric sequence, with $n$ terms,
    • The nth term is a<em>n=a</em>1rn−1a<em>n = a</em>1 r^{n-1}.

Geometric Series

  • Definition: A geometric series is the sum of the terms of a geometric sequence.
  • Finite sum of the first $n$ terms:
    • S<em>n=a</em>1(1−rn)1−r,r≠1S<em>n = \dfrac{a</em>1\left(1 - r^n\right)}{1 - r}, \quad r \neq 1
    • Here $Sn$ is the sum of the first $n$ terms, $a1$ is the first term, $n$ is the number of terms, and $r$ is the common ratio.
  • Infinite sum (geometric series) when $|r|<1$:
    • S<em>∞=a</em>11−rS<em>\infty = \dfrac{a</em>1}{1 - r}
    • Condition for convergence: $|r| < 1$.
  • Special case when $r = 1$:
    • The series becomes $a1 + a1 + \cdots$ and does not converge to a finite sum; partial sums are $Sn = n a1$.
  • Examples from the transcript:
    • Finite sum example with $a_1 = 3$, $r = 4$, $n = 7$:
    • S<em>7=a</em>11−r71−r=31−471−4=31−16384−3=16383.S<em>7 = a</em>1 \dfrac{1 - r^7}{1 - r} = 3 \dfrac{1 - 4^7}{1 - 4} = 3 \dfrac{1 - 16384}{-3} = 16383.
    • Finite sum example with $a_1 = 7$, $r = 2$, up to the term $14336$:
    • The sequence is $7, 14, 28, \dots, 14336$.
    • If there are $n$ terms, $Sn = a1 \dfrac{r^n - 1}{r - 1}$.
    • Here, $14336 = 7 \cdot 2^{n-1} \Rightarrow 2^{n-1} = 2048 \Rightarrow n-1 = 11 \Rightarrow n = 12$.
    • Thus, S12=7212−12−1=7⋅4095=28665.S_{12} = 7 \dfrac{2^{12} - 1}{2 - 1} = 7 \cdot 4095 = 28665.
    • Infinite-sum example (pendulum length):
    • First term $a_1 = 25\,\text{cm}$, common ratio $r = 0.95$ (length decreases by 5% each swing).
    • Since $|r| < 1$, the sum to infinity exists:
    • S<em>∞=a</em>11−r=251−0.95=250.05=500 cm.S<em>\infty = \dfrac{a</em>1}{1 - r} = \dfrac{25}{1 - 0.95} = \dfrac{25}{0.05} = 500\,\text{cm}.
    • Finite-sum exercise (Joey’s bamboo bank):
    • Weekly saving doubles: $a1 = 1$ peso, $r = 2$, after 10 weeks: S</em>10=a1r10−1r−1=210−12−1=1023.S</em>{10} = a_1 \dfrac{r^{10} - 1}{r - 1} = \dfrac{2^{10} - 1}{2 - 1} = 1023.
    • Infinite-sum case with $r = 1$ (case not convergent): if the sequence is $7,7,7,\dots$ then $Sn = 7n$ and there is no finite $S\infty$.
  • Important notes about the formulas:
    • For finite sums: S<em>n=a</em>1(1−rn)1−r,r≠1.S<em>n = \dfrac{a</em>1\left(1 - r^n\right)}{1 - r}, \quad r \neq 1.
    • For a convergent infinite sum: S\infty = \dfrac{a1}{1 - r}, \quad |r| < 1.
  • Real-world relevance and implications:
    • Geometric series model growth/decay processes with constant percentage change (e.g., savings with compound interest, depreciation, population growth with constant rate).
    • Convergence criteria are important: if $|r| \ge 1$, the infinite sum does not converge; only finite sums are meaningful in those cases.

Geometric Means

  • Purpose: Geometric means are the interior terms of a geometric progression when the first term and last term (extremes) are given.
  • Key definitions:
    • Extremes: the first term $a$ and the last term $b$ of a geometric sequence with $n$ terms.
    • Geometric means: the terms between the extremes; there are $n-2$ geometric means.
    • Common ratio: r=(ba)1n−1.r = \left(\dfrac{b}{a}\right)^{\frac{1}{n-1}}.
    • The means themselves are: ar, ar2, …, ar n−2.a r,\ a r^2,\ \dots,\ a r^{\,n-2}.
  • Notation used in the transcript:
    • $n$ = number of terms in the sequence
    • $k$ (mentioned as the number of interior terms minus one in some slides) relates to placement of means; the key takeaway is that there are $n-2$ interior geometric means.
  • Example demonstrations from the transcript:
    • Think of a number between 3 and 75: with two extremes 3 and 75, the geometric mean is
    • 3⋅75=225=15.\sqrt{3 \cdot 75} = \sqrt{225} = 15.
    • Think of three numbers between 1 and 16: for a three-term GP with ends 1 and 16, the middle term (the single geometric mean) is
    • 1⋅16=16=4.\sqrt{1 \cdot 16} = \sqrt{16} = 4.
    • Find two numbers between 125 and 64: for two interior means between 125 and 64 (i.e., a GP of three terms with ends 125 and 64), the common ratio is
    • r=(64125)12=64125=8125=855 (approx.),r = \left(\dfrac{64}{125}\right)^{\frac{1}{2}} = \sqrt{\dfrac{64}{125}} = \dfrac{8}{\sqrt{125}} = \dfrac{8}{5\sqrt{5}}\text{ (approx.)},
    • the geometric mean itself is 125⋅64=8000=405≈89.44.\sqrt{125 \cdot 64} = \sqrt{8000} = 40\sqrt{5} \approx 89.44.
    • Find the geometric mean between 27 and 243:
    • 27⋅243=6561=81.\sqrt{27 \cdot 243} = \sqrt{6561} = 81.
  • How to use GM to construct a GP:
    • Given first term $a$ and last term $b$ with $n$ terms, compute the common ratio using the formula above, then insert the interior terms as $a r, a r^2, \dots, a r^{n-2}$.
  • Connections to real-world ideas:
    • Geometric means arise naturally when interpolating in a GP, analogous to how arithmetic means interpolate in an AP.
    • Useful in finance and growth-decay models where proportional changes accumulate multiplicatively.
  • Quick recap of formulas:
    • Common ratio from endpoints: r=(ba)1n−1.r = \left(\dfrac{b}{a}\right)^{\frac{1}{n-1}}.
    • Interior terms (geometric means) lauded as ark,k=1,2,…,n−2.a r^k, \quad k = 1,2,\dots, n-2.
  • Ethical, philosophical, or practical implications:
    • The GM approach emphasizes proportional relationships; misinterpreting endpoints can lead to incorrect inferences about the middle terms.
    • In data interpolation or modeling, ensure the assumption of a constant ratio is appropriate for the context.
  • Summary connections to prior material:
    • Builds directly on the concept of a geometric sequence and extends to sums (series) and interpolations (means).