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November 21, 2024General Term of a Geometric Progression: When we say that a collection of objects is listed in a sequence, we usually mean that the collection is organised so that the first, second, third, and so on terms may be identified. An example of a sequence is the quantity of money deposited in a bank over a period of time. Progressions are sequences that follow specified patterns.
A geometric progression is a sequence in which each of its terms can be generated by multiplying or dividing the preceding term by a constant quantity. The fixed quantity is known as the common ratio. In this article, we will study the definition of geometric progression and the general term of a geometric progression and solve some examples.
A sequence is a group of numbers in which one number is identified as the first, another as the second, another as the third, etc.
Consider the following arrangement of numbers:
\(2, 4, 6, 8, 10, 12,……..\)
\(4, 8, 12, 16, 20,……\)
Numbers are organised in a specific sequence according to various principles in the above arrangements. The numbers are even natural numbers in the first arrangement, while in the second arrangement, the numbers are multiples of \(4.\) Each of the forms listed above is a sequence.
So, a sequence is an arrangement of numbers in a definite order according to some rule.
The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) etc. The position of the terms is denoted by the subscripts. \({a_1}\) is the first number or the number at first place in the sequence, the second number is called the second term and is denoted by \({a_2}\) and so on.
The \({n^{{\rm{th}}}}\) term is the number at the \({n^{{\rm{th}}}}\) position of the sequence and is denoted by \({a_n}.\) The \({n^{{\rm{th}}}}\) term is also called a general term of the sequence.
For example, \(7, 14, 21, 28,…….\) is a sequence whose first term is \(7,\) i.e., \({a_1} = 7,\) the second term is \(14,\) i.e., \({a_2} = 14,\) the third term is \(21,\) i.e., \({a_3} = 21,\) the fourth term is \(28,\) i.e., \({a_4} = 28,\) and so on.
A progression is a sequence in which the terms are written under fixed conditions. The three types of progression are arithmetic progression, geometric progression, and harmonic progression. Here we will study a particular type of progression called Geometric Progression (GP) or Geometric Sequence and the \({n^{{\rm{th}}}}\) term or a general term of the GP.
A sequence in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity is called a geometric progression. The fixed quantity is called the common ratio.
Consider the below examples:
(i) \(2, 4, 8, 16, 32,……\)
(ii) \(6, 2, 23,…….\)
In case (i), we have \({a_1} = 2,\frac{{{a_2}}}{{{a_1}}} = 2,\frac{{{a_3}}}{{{a_2}}} = 2,\frac{{{a_4}}}{{{a_3}}} = 2\) and so on.
In case (ii), we have \({a_1} = 6,\frac{{{a_2}}}{{{a_1}}} = \frac{2}{6} = \frac{1}{3},\frac{{{a_3}}}{{{a_2}}} = \frac{{\frac{2}{3}}}{2} = \frac{1}{3}\) and so on.
? Learn Concepts on Geometric Progression
It is observed that in each case, every term except the first term has a constant ratio to the term preceding it.
This constant ratio is called the common ratio, which is denoted by \(r.\)
Such sequences are called a geometric sequence or geometric progression, which is abbreviated as GP.
A sequence \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) is called GP, if each term is non-zero and
\(\frac{{{a_{k + 1}}}}{{{a_k}}} = r\) (constant), for \(k \ge 1\)
By substituting \({a_1} = a\) we obtain a geometric progression \(a,ar,a{r^2},a{r^3}, \ldots \ldots \) Where a is the first term, and \(r\) is called the common ratio of the GP.
Finite GP: A GP having a finite number of terms is called a finite GP. A finite GP has the last term.
Example: \(1, 3, 9, 27, 81, 243, 729.\)
Here, the first term, \({a_1} = 1\)
The common ratio, \(r = \frac{3}{1} = \frac{9}{3} = \frac{{27}}{9} = 3\)
Number of terms, \(n=7\)
Last term, \({a_n} = {a_7} = 729\)
Infinite GP: An infinite GP is a GP that does not have a fixed number of terms. In other words, a GP that has an infinite number of terms is called an infinite GP.
Example: \(3,\frac{3}{2},\frac{3}{4},\frac{3}{8}, \ldots \ldots \ldots \)
Here, the first term, \({a_1} = 3\)
The common difference, \(r = \frac{{\frac{3}{2}}}{3} = \frac{{\frac{3}{4}}}{{\frac{3}{2}}} = \frac{{\frac{3}{8}}}{{\frac{3}{4}}} = \frac{1}{2}\)
Number of terms, \(n = \infty,\) or we cannot define the number of terms
Last term, \({a_n}\) also cannot be found out.
Finding the \({n^{{\rm{th}}}}\) term or sum of \(n\) terms of a geometric progression with a large number of terms would be tricky if we didn’t have the formulae to find them. With these formulae, we will use the following notations:
\(a=\)the first term
\(r=\)the common ratio
\(l=\)the last term
\(n=\)the number of terms
Consider a GP with the first term as \(a\) and common ratio \(r.\) We know that the second term is obtained by multiplying \(a\) by \(r\) and the third by multiplying the second term by \(r.\)
So, \({a_2} = ar\) and \({a_3} = a{r^2}.\)
We can write few more terms as below:
First term\( = {a_1} = a = a{r^{1 – 1}}\)
Second term\( = {a_2} = ar = a{r^{2 – 1}}\)
Third term\( = {a_3} = a{r^2} = a{r^{3 – 1}}\)
Fourth term\( = {a_4} = a{r^3} = a{r^{4 – 1}}\)
Fifth term\( = {a_5} = a{r^4} = a{r^{5 – 1}}\)
We may see a pattern in the above terms.
Also, the \({15^{{\rm{th}}}}\) term can be written as \({a_{15}} = a{r^{14}} = a{r^{15 – 1}}\)
Therefore, the above pattern suggests that the \(n^{t h}\) term or a general term of a GP is given by
\({a_n} = a{r^{n – 1}}\)
Hence, a GP can be written as \(a,a{r^2},a{r^3},a{r^4}, \ldots \ldots a{r^{n – 1}}\) or \(a,a{r^2},a{r^3},a{r^4}, \ldots \ldots a{r^{n – 1}}, \ldots \ldots \) accordingly, as to whether the GP is finite or infinite.
Let \({x_1},{x_2},{x_3}, \ldots \ldots \ldots ,{x_n}\) is a given sequence. Thus the expression \({x_1} + {x_2} + {x_3} + \ldots \ldots \ldots , + {x_n}\) is called the series associated with the given sequence. The series is finite or infinite, depending on whether the given sequence is finite or infinite.
Let \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) be a given geometric sequence. Thus the expression \({a_1} + {a_2} + {a_3} + \ldots \ldots \ldots , + {a_n}\) is called the geometric series associated with the given sequence. The geometric series is finite or infinite, depending on whether the given sequence is finite or infinite.
If \({S_n}\) denote the sum of first \(n\) terms of the geometric series. Then,\({s_n} = \frac{{a\left( {1 – {r^n}} \right)}}{{1 – r}}\) or \({s_n} = \frac{{a\left( {{r^n} – 1} \right)}}{{r – 1}}\)
Q.1. In a GP, the third term is \(24,\) and the sixth term is \(192.\) Find the tenth term.
Ans: Given: \({a_3} = 24\) and \({a_6} = 192,{a_{10}} = \)?
\({a_3} = a{r^2} = 24 \ldots \ldots {\rm{(i)}}\) and \({a_6} = a{r^5} = 192 \ldots \ldots {\rm{(ii)}}\)
Dividing \({\rm{(ii)}}\) by \({\rm{(i)}},\) we get
\(\frac{{a{r^5}}}{{a{r^2}}} = \frac{{192}}{{24}}\)
\( \Rightarrow {r^3} = 8\)
\( \Rightarrow {r^3} = {2^3}\)
\( \Rightarrow r = 2\)
Substituting \(r=2\) in equation \({\rm{(i)}},\) we get
\(a{(2)^2} = 24\)
\( \Rightarrow 4a = 24\)
\( \Rightarrow a = 6\)
So, \({a_{10}} = a{r^9} = 6 \times {(2)^9}\)
\( \Rightarrow {a_{10}} = 3072\)
Therefore, the tenth term of the given GP is \(3072.\)
Q.2. Find the eighth term of the geometric progression \(5, 10, 20,…..\)
Ans: The given geometric progression is \(5, 10, 20,…..\)
Here, first term \(a=5,\) common ratio \(r = \frac{{10}}{5} = 2\)
We know that, \({n^{{\rm{th}}}}\) term of a geometric progression is \({a_n} = a{r^{n – 1}}\)
So, \({a_8} = a{r^{8 – 1}} = a{r^7} = 5 \times {(2)^7} = 5 \times 128 = 640\)
Therefore, the eighth term of the given geometric progression is \(640.\)
Q.3. Find the \({19^{{\rm{th}}}}\) term of the series \(\sqrt 3 + \frac{1}{{\sqrt 3 }} + \frac{1}{{3\sqrt 3 }} + \ldots \ldots \)
Ans: The given series is \(\sqrt 3 + \frac{1}{{\sqrt 3 }} + \frac{1}{{3\sqrt 3 }} + \ldots \ldots \)
Here, \(\frac{{{a_2}}}{{{a_1}}} = \frac{{\frac{1}{{\sqrt 3 }}}}{{\sqrt 3 }} = \frac{1}{3},\frac{{{a_3}}}{{{a_2}}} = \frac{{\frac{1}{{3\sqrt 3 }}}}{{\frac{1}{{\sqrt 3 }}}} = \frac{1}{3}\) and so on.
\( \Rightarrow \sqrt 3 + \frac{1}{{\sqrt 3 }} + \frac{1}{{3\sqrt 3 }} + \ldots \ldots \)is a geometric series with a common ratio \(\frac{1}{3}.\)
So, the first term \(=a=\sqrt{3}\) and the common ratio \(r = \frac{1}{3}.\)
We know that, \({n^{{\rm{th}}}}\) term of a geometric progression is \({a_n} = a{r^{n – 1}}\)
We have \({a_{19}} = a{r^{19 – 1}} = a{r^{18}} = \sqrt 3 \times {\left( {\frac{1}{3}} \right)^{18}}\)
\( \Rightarrow {a_{19}} = {\left( {\frac{{\sqrt 3 }}{3}} \right)^{18}}\)
Therefore, the \({19^{{\rm{th}}}}\) term of the given series is \({\left( {\frac{{\sqrt 3 }}{3}} \right)^{18}}\)
Q.4. Find the \({20^{{\rm{th}}}}\) term and \({n^{{\rm{th}}}}\) term of the GP \(\frac{5}{2},\frac{5}{4},\frac{5}{8}, \ldots \ldots \)
Answer: The given geometric progression is \(\frac{5}{2},\frac{5}{4},\frac{5}{8}, \ldots \ldots \)
Here, the first term \(a = \frac{5}{2},\) and the common ratio \(r = \frac{{{a_2}}}{{{a_1}}} = \frac{{\frac{5}{4}}}{{\frac{5}{2}}} = \frac{1}{2}\)
We know that, nth term of a geometric progression is \({a_n} = a{r^{n – 1}}\)
\({a_{20}} = a{r^{20 – 1}} = a{r^{19}} = \frac{5}{2} \times {\left( {\frac{1}{2}} \right)^{19}}\)
Therefore, \({a_{20}} = \frac{5}{2} \times {\left( {\frac{1}{2}} \right)^{19}}\)
Also, \({a_n} = \frac{5}{2} \times {\left( {\frac{1}{2}} \right)^{n – 1}}\)
So, the \({20^{{\rm{th}}}}\) term is \(\frac{5}{2} \times {\left( {\frac{1}{2}} \right)^{19}}\) and \({n^{{\rm{th}}}}\) term of the given GP is \(\frac{5}{2} \times {\left( {\frac{1}{2}} \right)^{n – 1}}.\)
Q.5. Find the \({12^{{\rm{th}}}}\) term of a GP whose \({8^{{\rm{th}}}}\) term is \(192,\) and the common ratio is 2.
Ans: Given \({8^{{\rm{th}}}}\) term is \(192 \Rightarrow {a_8} = a{r^{8 – 1}} = a{r^7} = 192\)
Also, the common ratio \(r=2\)
\({a_{12}} = ?\)
Now, \(a{(2)^7} = 192\)
\( \Rightarrow a = \frac{{192}}{{{2^7}}}\)
So, \({a_{12}} = a{r^{11}}\)
\( = \frac{{192}}{{{2^7}}} \times {2^{11}}\)
\( = 192 \times {2^4}\)
\( = 192 \times 16\)
\( = 3072\)
Therefore, the \({12^{{\rm{th}}}}\) term of the given geometric progression is \(3072.\)
In this article, we have studied the definition of sequence, progression and geometric progression. Also, we have studied to find the general term of the geometric progression and solved some example problems on the same.
Q.1. What is geometric progression?
Ans: A sequence in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity is called a geometric progression. The fixed quantity is called the common ratio.
In other words, a sequence \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) is called GP, if each term is non-zero and \(\frac{{{a_{k + 1}}}}{{{a_k}}} = r\) (constant), for \(k \ge 1\)
Q.2. What is the general term of a geometric progression (GP)?
Ans: The general term of a GP is given by \({a_n} = a{r^{n – 1}}.\)
Where, \(a=\)the first term, \(r=\)the common ratio and \(n=\)the number of terms.
Q.3. What is the common ratio in GP?
Ans: The common ratio is the number multiplied (or divided) at each level of a geometric sequence
Q.4. What is the formula to find the general term of geometric progression?
Ans: The formula to find the general term of a GP is given by \({a_n} = a{r^{n – 1}}\)
Where, \(a=\)the first term, \(r=\)the common ratio and \(n=\)the number of terms.
Q.5. What is geometric series?
Ans: Let \({a_1},{a_2},{a_3}, \ldots \ldots \ldots ,{a_n}\) an be a given geometric sequence. Thus the expression \({a_1} + {a_2} + {a_3} + \ldots \ldots \ldots , + {a_n}\) is called the geometric series associated with the given sequence. The series is finite or infinite, according to as the given sequence is finite or infinite.
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