Geometric progression – Introduction:
If in a sequence of terms each term is constant multiple of the preceding term, then theÂ sequence is called a geometric progression.(G.P), whereas the constant multiplier is called the common ratio.
In an A.P, the difference between \(n^{th}\) term and \((n-1)^{th}\) term will be a constant which is known as the common difference of the A.P.
An Arithmetic Progression is in the form,
\(a, a+d, a+2d, a+3d â€¦.. a + (n-1)d\)
Now, what if the ratio of \(n^{th}\) term to \((n-1)^{th}\) term in a sequence is a constant?
For example, consider the following sequence,
\(2, 4, 8, 16, 32,â€¦â€¦..\)
You can see that,
\(\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} = 2\)
Similarly,
Consider a series \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},â€¦â€¦â€¦.\)
\(\frac{\frac{1}{2}}{1}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\large \frac{\frac{1}{8}}{\frac{1}{4}} = \frac{\frac{1}{16}}{\frac{1}{8}} = \frac{1}{2}\)
In the given examples, the ratio is a constant. Such sequences are called Geometric Progressions. It is abbreviated as G.P.
A sequence \(a_1,a_2,a_3,â€¦â€¦.a_n,â€¦.\) is a G.P, then \(\frac{a_{k+1}}{a_k} = r~~~~~~~~~~~~ [k>1]\)
Where r is a constant which is known as common ratio and none of the terms in the sequence is zero.
General term of a Geometric Progression:
We had learned to find the \(n^{th}\) term of an A.P, which was
\(a_n \)= \( a + (n-1)d\)
Similarly, in case of G.P,
Let a be the first term and r be the common ratio,
Then the second term, \(a_2 = a \times r = ar \)
Third term, \(a_{3} = a_{2} \times r = ar \times r = ar^{2}\)
Similarly, \(n^{th}\) term, \(a_{n} \) = \(ar^{n-1}\)
General term of a Geometric Progression
Â an = ar^{nâˆ’1} |
Common Term
Consider the sequence a, ar, ar^{2}, ar^{3},……
First term = a
Second term = ar
Third term = ar^{2}
Similarly nth term, t_{n} =Â ar^{n-1}
Thus, Common ratio = (Any term) / (Preceding term)
= t_{n} / t_{n-1}
= (ar^{n – 1} ) /(ar^{n – 2})
= r
Thus, the general term of a G.P is given byÂ ar^{n-1} and the general form of a G.P is a + ar + ar^{2} Â + â€¦..
For Example: r = t_{2} / t_{1} = ar / a = r
Finite and Infinite Geometric Progression
The terms of a finite G.P. can be written as \(a, ar, ar^2, ar^3,â€¦â€¦ar^{n-1}\)
Terms of an infinite G.P. can be written as \(a, ar, ar^{2}, ar^{3}, â€¦â€¦ar^{n-1},â€¦â€¦.\)
\(a + ar + ar^2 + ar^3 + â‹¯ + ar^{n-1}\) is called finite geometric series.
\( a + ar + ar^2 + ar^3 + â‹¯ + ar^n + â‹¯ \) is called infinite geometric series.
Sum of n terms of a G.P:
Consider the G.P,
\(a,ar,ar^2,â€¦..ar^{n-1}\)
Let \(S_n,a,r\) be the sum of n terms, first term and ratio of the G.P respectively.
Then, \(S_n\) = \(a + ar + ar^2 + â‹¯ + ar^{n-1}\) —(1)
There are two cases, either \(r = 1\) or \(r â‰ 1\)
If r=1, then \(S_n\) = \( a + a + a + â‹¯ a\) = \(na\)
When \(r â‰ 1\),
Multiply (1) with r gives,
\(rS_n\) = \( ar + ar^2 + ar^3 + â‹¯ + ar^{n-1} + ar^n\)—(2)
Subtracting (1) from (2) gives
\(rS_n – S_n = (ar + ar^{2} + ar^{3} + …. + ar^{n-2} + ar^{n-1} + ar^{n}) – (a + ar + ar^{2} + …. + ar^{n-2} + ar^{n-1})\)
\((r – 1) S_n = ar^{n} – a = a(r^{n}-1)\)
\(S_n = a\frac{(r^{n}-1)}{(r – 1)} = a\frac{(1 – r^{n})}{(1 – r )}\)
Sum of n termsÂ
Sn = \(\frac{a(r^n – 1)}{r-1}\); Where r \(\neq\) 1 |
Example Problems
Example 1: If \(n^{th}\) term of the G.P 3, 6, 12, â€¦. is 192, then what is the value of n?
Solution: First, we have to find the common ratio
\(r\)= \(\frac{6}{3}\) = \(2\)
Since the first term, \(a\) = \(3\)
\(a_n\) = \(ar^{n-1}\)
\(192\) = \(3 \times 2^{n-1}\)
\(2^{n-1} = \frac{192}{3} = 64 = 2^6\)
\(n – 1 = 6 \)
n = 7
Therefore, 192 is \(7^{th}\) term of the G.P.
Example 2: \(5^{th}\) term and \(3^{rd}\) term of a G.P is 256 and 16 respectively. Find its \(8^{th}\) term.
Solution: \(ar^4\) = \(256\)—(1)
\(ar^2\) = \(16\)—(2)
Dividing (1) by (2) gives,
\(\frac{ar^4}{ar^2}\) = \(\frac{256}{16}\)
\(r^2\) = \(16\)
\(r\)= \(4\)
Substituting \(r\) = \(4\) in (2) gives,
\(aÃ—4^2\) = \(16\), \(a\)= \(1\)
\(a_8\) = \(ar^7\)
=\( 1Ã—4^7\) = \(16384\)
Example 3: Find the sum of 6 terms of the G.P 4, 12, 36,â€¦..
Solution: \(a\) = \(4\)
Common ratio,\(r = \frac{12}{4} = 3\)
\(n\) = \(6\)
Sum of n terms of a G.P,
\(S_n\) = \(\frac{a(r^n-1)}{(r-1)}\)
\(S_6\) = \(\frac{4(3^6-1)}{(3-1)}\)
=\(\frac{4(729-1)}{(2)}\) = \(2 Ã— 728 \) = \(1456\)
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