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

• Definition of Geometric Progression
• General Form and General Term of a Geometric Progression
• Sum of n terms of a Geometric Progression
• Definition of Geometric Mean
• Position of a term in a Geometric Progression
• Selection of Terms in Geometric Progression
• Sum of an infinite Geometric Progression
• Geometric Progression Formulae
• Properties of Geometric Progression
• Relation between Arithmetic Means and Geometric Means
• Problems on Geometric Progression

11 and 12 Grade Math 

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