# 27. Geometric Progressions¶

A succession of numbers is said to be in geometric progression or geometric sequence if the ratio of any term and the term preceding it is constant throughout. This constant is called common ratio of the G. P.

Example: $$1, 2, 4, 8, 16, ...$$

Here, $$\frac{\text{2nd term}}{\text{1st term}} = \frac{\text{3rd term}}{\text{2nd term}} = ... = 2$$.

Also, $$1, 3, 9, 27, ...$$ are in geometric progression whose first term is $$1$$ and common ratio is $$3$$.

Also, $$2, -4, 8, -16, ...$$ are in geometric progression whose first term is $$2$$ and common ratio is $$-2$$.

## 27.1. Properties of a G. P.¶

1. If the each term of a G. P. be multiplied by a non-zero number, then the sequence obtained is also a G. P.

Proof: Let the given G. P. be $$a, ar, ar^2, ar^3, ...$$

Let $$k$$ be a non-zero number, the sequence obtained by multiplying each term of the given G. P. by $$k$$ is $$ak, ark, ar^2k, ar^3k, ...$$

Obviously, the series is in G. P. with the same constant ratio as previous G. P. i.e. $$r$$.

Again, dividing each term of G. P. $$a, ar, ar^2, ar^3, ...$$ we obtain the sequence $$\frac{a}{k}, \frac{ar}{l^2}, \frac{ar^2}{k^3}, \frac{ar^3}{k4}, ...$$

It is evident that this sequence is a G. P. whose common ratio is $$r$$.

2. The reciprocals of the terms of a G. P. are also in G. P.

Proof: Let the G.P. be $$a, ar, ar^2, ar^3, ...$$ the sequence whose terms are reciprocal of this G. P. is $$\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, ...$$

It is clear that this sequence is in G. P. whose first term is $$\frac{1}{a}$$ and common ratio is $$\frac{1}{r}$$.

## 27.2. Sum of the First n Terms of a G. P.¶

Let $$a$$ be the first term and $$r$$ be the common ratio of a G. P. and $$S_n$$ be the sum of its first $$n$$ terms.

Case I: When $$r\neq 1$$

Now, $$Sn = ~a + ar + ar^2 + ... + ar^{n - 2} + ar^{n - 1}$$ and $$4S_n = ~~~~ar + ar^2 + ... + ar^{n - 2} + ar^{n - 1} + ar^n$$

Subtracting bottom one from top one we get

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

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

Case II. When $$r = 1$$

$$S_n = a + a + ... + a = na$$ and G. P. is also an A. P. with common difference of $$0$$.

## 27.3. Sum of Infinite Terms of a G. P., when $$|r| < 1$$¶

Proceeding similarly as previous section

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

However, when $$|r| < 1$$, then $$r^n$$ approaches zero as $$n$$ approaches $$\pm \infty$$. Thus,

$$S_{\infty} = \frac{a}{1 - r}, ~~~ (|r| < 1)$$

## 27.4. Recurring Decimals¶

Recurring decimals are a very interesting and nice example to demonstrate the infinite G. P. and the value can be obtained by the formula derived in previous section. Consider a recurring decimal $$.\dot{7}$$. Now,

$.\dot{7} = .777777 ... \text{to }\infty$
$.\dot{7} = .7 + .07 + .007 + .0007 + ...$
$.\dot{7} = \frac{7}{10} + \frac{7}{100} + \frac{7}{1000} + ...$
$.\dot{7} = \frac{7}{10} + \frac{7}{10^2} + \frac{7}{10^3} + ...$
$.\dot{7} = 7\left(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...\right)$
$.\dot{7} = \frac{7}{9}$

We can find this using an alternative method. Let $$S = .\dot{7}$$. Multiplying it with 10, we have $$10S = 7.\dot{7}$$.

Subtracting we get $$9S = 7$$ and thus we obtain the same result.

## 27.5. Geometric Mean¶

Like arithmetic means we also have geometric means. Say two numbers $$a$$ and $$b$$ are in G. P. and $$x$$ is a geometric mean between them then by definition

$\frac{x}{a} = \frac{b}{x}$

$$\Rightarrow x^2 = ab \Rightarrow x = \sqrt{ab}$$

If $$G_1, G2, ..., G-n$$ are $$n$$ geometric means between two numbers $$a$$ and $$b$$, then

$$G1G_2...G_n = \sqrt{ab}^n = G^n$$.

Proof: $$b$$ is $$(n + 2)nd$$ term. Thus, $$b = ar^{n + 1}$$ where common ratio is $$r$$.

Thus, $$G_1 = ar, G_2 = ar^2, ..., Gn = ar^n$$

$G_1G_2...G_n = = ar^{1 + 2 + ... + n}$
$G_1G_2...G_n = = ar^\frac{n(n + 1)}{2} = \sqrt{ab}^n$

If $$a_1, a_2, ..., a_n$$ are $$n$$ positive numbers in G. P. then their geometric mean is given by

$$G = (a_1a_2 ... a_n)\frac{1}{n}$$

## 27.7. Arithmetico Geometric Series¶

If the terms of an A. P. are multiplied by the corresponding terms of a G. P., then the new series obtained is called an Arithmetico Geometric series.

Example: If the terms of the arithmetic series $$2 + 5 + 8 + 11 + ...$$ are multiplied with the corresponding terms of the geometric series $$x + x^2 + x^3 + ...$$ then the following arithmetico-geometric series is formed

$$2x + 5x^2 + 8x^3 + 11x^4 + ...$$

## 27.8. Sum of $$n$$ terms of an Arithmetico-Geometric Series¶

Let $$a_1, a_2, ..., a_n$$ be an A. P. and $$b_1, b_2, ..., b_n$$ be a G. P. Let $$d$$ be the common difference of the A. P. and $$r$$ be the common ratio of the G. P.

Let

$S_n = ab + (a + d)br + (a + 2d)br^2 + ... + [a + (n - 1)d]br^{n - 1}$

We multiply each term by $$r$$ and write first term below second, second term below the third and so on.

$rS_n = abr + (a + d)br^2 + (a + 2d)br^3 + ... + [a + (n - 1)d]br^n$

Subtracting we get

$(1 - r)S_n = ab + dbr + dbr^2 + ... + dbr^{n - 1} - [a + (n - 1)d]br^n$
$= ab + \frac{dbr(1 - r^{n - 1})}{1 - r} - [a + (n - 1)d]br^n$
$S_n = \frac{ab}{1 - r} + \frac{dbr(1 - r^{n - 1})}{(1 - r)^2} - \frac{[a + (n - 1)d]br^n}{1 - r}~~~~(r \neq 1)$

If $$-1<r<1$$ then

$\lim_{n\rightarrow \infty}r^n = 0 \Rightarrow \lim_{n\rightarrow \infty}nr^n = 0$

Therefore, sum of an infinite number of terms of an arithmetico-geometric series is given by

$S_{\infty} = \frac{ab}{1 - r} + \frac{dbr}{(1 - r)^2}$