Please explain geometric and harmonic progressions?

Feb 5, 2017

Arithmetic progression: ${a}_{n} = A + n d$

Geometric progression: ${a}_{n} = A {n}^{q}$

Harmonic progression: ${a}_{n} = \frac{1}{A + n h}$

Explanation:

We also have to introduce the arithmetic progression, since the definition of the harmonic progression requires it.

Arithmetic progression

An arithmetic progression is a sequence of numbers:

${a}_{1} , {a}_{2} , \ldots , {a}_{n} , \ldots$

such that the difference between two consecutive numbers is constant:

${a}_{n} - {a}_{n - 1} = d$ for every $n$

If we define:

$A = {a}_{1} - d$

then we have:

${a}_{1} = A + d$

${a}_{2} - {a}_{1} = d \implies {a}_{2} = A + 2 d$

and so on, so we can see that the terms of an arithmetic progression can be expressed in the form:

${a}_{n} = A + n d$

If we consider three consecutive terms we have:

$\frac{{a}_{n - 1} + {a}_{n + 1}}{2} = \frac{\left({a}_{n} - d\right) + \left({a}_{n} + d\right)}{2} = \frac{2 {a}_{n}}{2} = {a}_{n}$

so each term is the arithmetic mean of the terms adjacent to it.

Geometric progression

A geometric progression is a sequence of numbers:

${a}_{1} , {a}_{2} , \ldots , {a}_{n} , \ldots$

such that the ratio between two consecutive numbers is constant:

${a}_{n} / {a}_{n - 1} = q$ for every $n$ and with $q \ne 0$

If we define:

$A = {a}_{1} / q$

then we have:

${a}_{1} = A q$

${a}_{2} / {a}_{1} = q \implies {a}_{2} = A {q}^{2}$

and so on, so we can see that the terms of an arithmetic progression can be expressed in the form:

${a}_{n} = A {q}^{n}$

If $q > 0$ and we consider three consecutive terms we have:

$\sqrt{{a}_{n - 1} {a}_{n + 1}} = \sqrt{\left({a}_{n} / q\right) \left({a}_{n} \cdot q\right)} = \sqrt{{a}_{n}^{2}} = {a}_{n}$

so each term is the geometric mean of the terms adjacent to it.

Harmonic progression

A harmonic progression is a sequence of numbers:

${a}_{1} , {a}_{2} , \ldots , {a}_{n} , \ldots$

such that their reciprocal constitute an arithmetic progression

$\frac{1}{a} _ n - \frac{1}{a} _ \left(n - 1\right) = h$ for every $n$

If we define:

$A = \frac{1}{a} _ 1 - h$

then we have:

${a}_{1} = \frac{1}{A + h}$

$\frac{1}{a} _ 2 - \frac{1}{a} _ 1 = h \implies \frac{1}{a} _ 2 = A + 2 h$

and so on, so we can see that the terms of an arithmetic progression can be expressed in the form:

${a}_{n} = \frac{1}{A + n h}$

If we consider three consecutive terms we have:

$\frac{2}{\frac{1}{a} _ \left(n + 1\right) + \frac{1}{{a}_{n - 1}}} = \frac{2}{\left(\frac{1}{a} _ n + h\right) + \left(\frac{1}{a} _ n - h\right)} = \frac{2}{\frac{2}{a} _ n} = {a}_{n}$

so each term is the harmonic mean of the terms adjacent to it.