# Geometric Progression And Sum Of GP

## Geometric progression – Introduction:

In an A.P, the difference between $$n^{th}$$ term and $$(n-1)^{th}$$ term will be a constant which is known as the common difference of the A.P.

An Arithmetic Progression is in the form,

$$a, a+d, a+2d, a+3d ….. a + (n-1)d$$

Now, what if the ratio of $$n^{th}$$ term to $$(n-1)^{th}$$ term in a sequence is a constant?

For example, consider the following sequence,

$$2, 4, 8, 16, 32,……..$$

You can see that,

$$\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2$$

Similarly,

Cosider a series $$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},……….$$

$$\frac{\frac{1}{2}}{1}$$ = $$\frac{\frac{1}{4}}{\frac{1}{2}}$$ = $$\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}$$

In the given examples, the ratio is a constant. Such sequences are called Geometric Progressions. It is abbreviated as G.P.

A sequence $$a_1,a_2,a_3,…….a_n,….$$ is a G.P, then $$\frac{a_{k+1}}{a_k} = r~~~~~~~~~~~~ [k>1]$$

Where r is a constant which is known as common ratio and none of the terms in the sequence is zero.

## General term of a Geometric Progression:

We had learned to find the $$n^{th}$$ term of an A.P, which was

$$a_n$$= $$a + (n-1)d$$

Similarly, in case of G.P,

Let a be the first term and r be the common ratio,

Then the second term, $$a_2 = a \times r = ar$$

Third term, $$a_{3} = a_{2} \times r = ar \times r = ar^{2}$$

Similarly, $$n^{th}$$ term, $$a_{n}$$ = $$ar^{n-1}$$

The terms of a finite G.P can be written as $$a, ar, ar^2, ar^3,……ar^{n-1}$$

Terms of an infinite G.P can be written as $$a, ar, ar^{2}, ar^{3}, ……ar^{n-1},…….$$

$$a + ar + ar^2 + ar^3 + ⋯ + ar^n$$ is called finite geometric series.

$$a + ar + ar^2 + ar^3 + ⋯ + ar^n + ⋯$$ is called infinite geometric series.

Example: If $$n^{th}$$ term of the G.P 3, 6, 12, …. is 192, then what is the value of n?

Solution: First, we have to find the common ratio

$$r$$= $$\frac{6}{3}$$ = $$2$$

Since the first term, $$a$$ = $$3$$

$$a_n$$ = $$ar^{n-1}$$

$$192$$ = $$3 \times 2^{n-1}$$

$$2^{n-1} = \frac{192}{3} = 64 = 2^6$$

$$n – 1 = 6$$

n = 7

Therefore, 192 is $$7^{th}$$ term of the G.P.

Example: $$5^{th}$$ term and $$3^{rd}$$ term of a G.P is 256 and 16 respectively. Find its $$8^{th}$$ term.

Solution: $$ar^4$$ = $$256$$—(1)

$$ar^2$$ = $$16$$—(2)

Dividing (1) by (2) gives,

$$\frac{ar^4}{ar^2}$$ = $$\frac{256}{16}$$

$$r^2$$ = $$16$$

$$r$$= $$4$$

Substituting $$r$$ = $$4$$ in (2) gives,

$$a×4^2$$ = $$16$$, $$a$$= $$1$$

$$a_8$$ = $$ar^7$$

=$$1×4^7$$ = $$16384$$

Sum of n terms of a G.P:

Consider the G.P,

$$a,ar,ar^2,…..ar^{n-1}$$

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

Then, $$S_n$$ = $$a + ar + ar^2 + ⋯ + ar^{n-1}$$ —(1)

There are two cases, either $$r = 1$$ or $$r ≠ 1$$

If r=1, then $$S_n$$ = $$a + a + a + ⋯ a$$ = $$na$$

When $$r ≠ 1$$,

Multiply (1) with r gives,

$$rS_n$$ = $$ar + ar^2 + ar^3 + ⋯ + ar^{n-1} + ar^n$$—(2)

Subtracting (1) from (2) gives

$$rS_n – S_n = (ar + ar^{2} + ar^{3} + …. + ar^{n-2} + ar^{n-1} + ar^{n}) – (a + ar + ar^{2} + …. + ar^{n-2} + ar^{n-1})$$

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

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

Example: Find the sum of 6 terms of the G.P 4, 12, 36,…..

Solution: $$a$$ = $$4$$

Common ratio,$$r = \frac{12}{4} = 3$$

$$n$$ = $$6$$

Sum of n terms of a G.P,

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

$$S_6$$ = $$\frac{4(3^6-1)}{(3-1)}$$

=$$\frac{4(729-1)}{(2)}$$ = $$2 × 728$$ = $$1456$$<