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**

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