In this section, we are going to learn about geometric progression.
What is geometric progression ?
A Geometric Progression is a sequence in which each term is obtained by multiplying a fixed non-zero number to the preceding term except the first term.
The fixed number is called common ratio. The common ratio is usually denoted by r.
General form of geometric progression :
The numbers of the form
a, ar, ar^{2}, ...ar^{n-1}
is called a Geometric Progression.
The number ‘a’ is called the first term and number ‘r’ is called the common ratio.
To find the common ratio, we use the formula
r = a_{2}/a_{1}
General term of the geometric progression :
t_{n} = a r^{(n-1)}
Example 1 :
Find out which of the following sequences are geometric sequences . For those geometric sequences, find the common ratio.
(i) 0.12, 0.24, 0.48,.........
Solution :
t_{1} = 0.12, t_{2} = 0.24 and t_{3} = 0.48
r = t_{2}/t_{1} r = 0.24 / 0.12 r = 24 / 12 r = 2 ----(1) |
r = t_{3}/t_{2} r = 0.48 / 0.24 r = 48/24 r = 2 ----(2) |
Since the common ratios are same, the given sequence is geometric progression.
The required common ratio is 2.
(ii) 0.004, 0.02, 1, ..........
Solution :
t_{1} = 0.004, t_{2} = 0.02 and t_{3} = 1
r = t_{2}/t_{1} r = 0.02 / 0.004 r = 20 / 4 r = 5 ----(1) |
r = t_{3}/t_{2} r = 1 / 0.02 r = 100/2 r = 50 ----(2) |
Since the common ratios are not same, the given sequence is not a geometric progression.
The required common ratio is 5.
Example 2 :
Find the 10^{th} term and the common ratio of the geometric sequence
1/4, -1/2, 1, -2,............
Solution :
To find the 10^{th} terms of the G.P we use the formula given below.
t_{n} = a r^{(n-1)}
a = 1/4, r = (-1/2) / (1/4) ==> -2 and n = 10
t_{10} = (1/4) (-2)^{(10 - 1)}
t_{10} = (1/4) (-2)^{9}
t_{10} = (1/4) (-512)
t_{10} = -128
Therefore the 10^{th} and common ratio of the given geometric sequence are -128 and -2 respectively.
After having gone through the stuff given above, we hope that the students would have understood about geometric progression.
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