Geometric Progression

I. Introduction

A sequence of numbers is said to be in $Geometric\ Progression$, if the ratio of any two consecutive terms is always constant.

For example, the series:

$5,\ 10,\ 20,\ 40...$

Here the ratio of any term to its previous term is always $2$.

II. Definitions & Common Notations

The first term is commonly denoted by $a$.

In our example, $a = 5$

The constant ratio of any term to the previous term is called the $common\ ratio$, and is commonly denoted using $r$.

In our example, $r = 2$

The $n\xasuper{th}$ term is commonly denoted by $T_n$

In our example, $T_4 = 40$

The sum of the first $n$ terms is commonly denoted by $S_n$.

In our example, $S_3 = 5 + 10 + 20 = 35$

III. Terms Of A Geometric Progression

In this section we will see how to compute the $n\xasuper{th}$ of a geometric progression, given the values of $a$ and $r$.

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$T_1 = a$

$T_2 = ar$

$T_3 = T_2 \times r = ar \times r = ar^2$

$T_4 = T_3 \times r = ar^2 \times r = ar^3$

$...$

$T_n = ar^{n - 1}$

IV. Sum Of Terms

Let us derive the formula for the sum of the first $n$ terms of this series.

$S_n = T_1 + T_2 + T_3 + ... + T_n$ $... eqn\ (i)$

Multiplying both sides by $r$ we get:

$r \times S_n = r \times (T_1 + T_2 + T_3 + ... + T_n)$

$r \times S_n = T_1\times r + T_2\times r + T_3\times r + ... + T_n\times r$

We know that any term of a geometric progression is obtained by multiplying the previous term by $r$

Therefore,

$T_2 = T_1 \times r$

$T_3 = T_2 \times r$

$T_n = T_{n-1} \times r$

Therefore we get:

$r \times S_n = T_2 + T_3 + T_4 + ... + T_{n+1}$ $... eqn\ (ii)$

Subtracting $equation\ (i)$ from $equation\ (ii)$, we get:

$r \times S_n - S_n = (T_2 + T_3 + T_4 + ... + T_{n+1}) - (T_1 + T_2 + T_3 + ... + T_n)$

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We can see that all terms except $T_{n + 1}$ from $eqn\ (ii)$ and $T_1$ from $eqn\ (i)$ will cancel out.

$\therefore S_n(r - 1) = T_{n + 1} - T_1$

$\Rightarrow S_n(r - 1) = ar^n - a = a(r^n - 1)$

$\Rightarrow S_n = \dfrac{a(r^n - 1)}{(r - 1)}$

V. Infinite GP Series

This is a special case of GP series where each term is smaller in magnitude than the previous term, that is $|T_{n+1}| < |T_n|$ for all $n$. As the terms of this series becomes progessively smaller in magnitude, their sum is within some finite limit. These types of series are called converging series and it is possible to find their sum.

As it can be seen, that since each term is smaller in magnitude than the previous term, the common ratio $r$ has to be less than $1$ in magnitude, that is: $0 \lt |r| \lt 1$.

Therefore, as $n$ grows larger and larger, $r^n$ becomes smaller and smaller and becomes almost $0$.

Therefore, the summation expression from our previous section becomes:

$S_n = \dfrac{a(r^n - 1)}{(r - 1)}$

$\Rightarrow S_n = \dfrac{a(1 - r^n)}{(1 - r)}$

When $n$ is very large, that is $n$ gets closer to infinity, $r^n$ becomes almost $0$ and can be ignored.

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Therefore, our summation formula becomes:

$S_n = \dfrac{a}{(1 - r)}$