# General Form and General Term of a Geometric Progression

We will discuss here about the general form and general term of a Geometric Progression.

The general form of a Geometric Progression is {a, ar, ar$$^{2}$$, ar$$^{3}$$, ar$$^{4}$$, ............}, where ‘a’ and ‘r’ are called the first term and common ratio (abbreviated as C.R.) of the Geometric Progression.

The nth or general term of a Geometric Progression

To prove that the general term or nth term of a Geometric Progression with first term ‘a’ and common ratio ‘r’ is given by t$$_{n}$$ = a ∙ r$$^{n - 1}$$

Proof:

Let us assume that t$$_{1}$$, t$$_{2}$$, t$$_{3}$$, t$$_{4}$$, ................., t$$_{n}$$, ............... be the given Geometric Progression with common ratio r. Then t$$_{1}$$ = a ⇒ t$$_{1}$$ = ar$$^{1 - 1}$$

Since t$$_{1}$$, t$$_{2}$$, t$$_{3}$$, t$$_{4}$$, ................., t$$_{n}$$, ............... is a Geometric Progression with common ratio r, therefore

$$\frac{t_{2}}{t_{1}}$$ = r ⇒ t$$_{2}$$ = t$$_{1}$$r ⇒ t$$_{2}$$ = ar ⇒ t$$_{2}$$ = ar$$^{2 - 1}$$

$$\frac{t_{3}}{t_{2}}$$ = r ⇒ t$$_{3}$$ = t$$_{2}$$r ⇒ t$$_{3}$$ = (ar)r ⇒ t$$_{3}$$ = ar$$^{2}$$ = t$$_{3}$$ = ar$$^{3 - 1}$$

$$\frac{t_{4}}{t_{3}}$$ = r ⇒ t$$_{4}$$ = t$$_{3}$$r ⇒ t$$_{4}$$ = (ar$$^{2}$$)r ⇒ t$$_{4}$$ = ar$$^{3}$$ = t$$_{4}$$ = ar$$^{4 - 1}$$

$$\frac{t_{5}}{t_{4}}$$ = r ⇒ t$$_{5}$$ = t$$_{4}$$r ⇒ t$$_{5}$$ = (ar$$^{3}$$)r ⇒ t$$_{5}$$ = ar$$^{4}$$ = t$$_{5}$$ = ar$$^{5 - 1}$$

Therefore, in general, we have, t$$_{n}$$ = ar$$^{n - 1}$$.

Alternate method to find the nth term of a Geometric Progression:

To find the nth term or general term of a Geometric Progression, let us assume that a, ar, ar$$^{2}$$, ar$$^{3}$$, a$$^{4}$$, .......... be the given Geometric Progression, where ‘a’ is  the first term and ‘r’ is the common ratio.

Now form the Geometric Progression a, ar, ar$$^{2}$$, ar$$^{3}$$, a$$^{4}$$, ......... we have,

Second term = a ∙ r = a ∙ r$$^{2 - 1}$$ = First term × (Common ratio)$$^{2 - 1}$$

Third term = a ∙ r$$^{2}$$ = a ∙ r$$^{3 - 1}$$ = First term × (Common ratio)$$^{3 - 1}$$

Fourth term = a ∙ r$$^{3}$$ = a ∙ r$$^{4 - 1}$$= First term × (Common ratio)$$^{4 - 1}$$

Fifth term = a ∙ r$$^{4}$$ = a ∙ r$$^{5 - 1}$$ = First term × (Common ratio)$$^{5 - 1}$$

Continuing in this manner, we get

nth term = First term × (Common ratio)$$^{n - 1}$$ = a ∙ r$$^{n - 1}$$

⇒ t$$_{n}$$ = a ∙ r$$^{n - 1}$$, [t$$_{n}$$ = nth term of the G.P. {a, ar, ar$$^{2}$$, ar$$^{3}$$, ar$$^{4}$$, ..........}]

Therefore, the nth term of the Geometric Progression {a, ar, ar$$^{2}$$, ar$$^{3}$$, .......} is t$$_{n}$$ = a ∙ r$$^{n - 1}$$

Notes:

(i) From the above discussion we understand that if ‘a’ and ‘r’ are the first term and common ratio of a Geometric Progression respectively, then the Geometric Progression can be written as

a, ar, ar$$^{2}$$, ar$$^{3}$$, ar$$^{4}$$, ................, ar$$^{n - 1}$$ as it is finite

or,

ar, ar$$^{2}$$, ar$$^{3}$$, ar$$^{4}$$, ................, ar$$^{n - 1}$$, ................ as it is infinite.

(ii) If first term and common ratio of a Geometric Progression are given, then we can determine its any term.

How to find the nth term from the end of a finite Geometric Progression?

Prove that if ‘a’ and ‘r’ are the first term and common ratio of a finite Geometric Progression respectively consisting of m terms then, the nth term from the end is ar$$^{m - n}$$.

Proof:

The Geometric Progression consists of m terms.

Therefore, nth term from the end of the Geometric Progression = (m - n + 1)th term from the beginning of the Geometric Progression = ar$$^{m - n}$$

Prove that if ‘l’ and ‘r’ are the last term and common ratio of a Geometric Progression respectively then, the nth term from the end is l($$\frac{1}{r}$$)$$^{n - 1}$$.

Proof:

From the last term when we move towards the beginning of a Geometric Progression we find that the progression is a Geometric Progression with common ratio 1/r. Therefore, the nth term from the end = l($$\frac{1}{r}$$)$$^{n - 1}$$.

Solved examples on general term of a Geometric Progression

1. Find the 15th term of the Geometric Progression {3, 12, 48, 192, 768, ..............}.

Solution:

The given Geometric Progression is {3, 12, 48, 192, 768, ..............}.

For the given Geometric Progression we have,

First term of the Geometric Progression = a = 3

Common ratio of the Geometric Progression = r = $$\frac{12}{3}$$ = 4.

Therefore, the required 15th term = t$$_{15}$$ = a ∙ r$$^{n - 1}$$ = 3 ∙ 4$$^{15 - 1}$$ = 3 ∙ 4$$^{14}$$ = 805306368.

2. Find the 10th term and the general term of the progression {$$\frac{1}{4}$$, -$$\frac{1}{2}$$, 1, -2, ...............}.

Solution:

The given Geometric Progression is {$$\frac{1}{4}$$, -$$\frac{1}{2}$$, 1, -2, ...............}.

For the given Geometric Progression we have,

First term of the Geometric Progression = a = $$\frac{1}{4}$$

Common ratio of the Geometric Progression = r = $$\frac{\frac{-1}{2}}{\frac{1}{4}}$$ = -2.

Therefore, the required 10th term = t$$_{10}$$ = ar$$^{10 - 1}$$ = $$\frac{1}{4}$$(-2)$$^{9}$$ = -128, and, general term, t$$_{n}$$ = ar$$^{n - 1}$$ = $$\frac{1}{4}$$(-2)$$^{n - 1}$$ = (-1)$$^{n - 1}$$2$$^{n - 3}$$

Geometric Progression

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