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
11 and 12 Grade Math
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