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We will learn how to find the position of a term in a Geometric Progression.
On finding the position of a given term in a given Geometric Progression
We need to use the formula of nth or general term of a Geometric Progression tn = arn−1.
1. Is 6144 a term of the Geometric Progression {3, 6, 12, 24, 48, 96, .............}?
Solution:
The given Geometric Progression is {3, 6, 12, 24, 48, 96, .............}
The first terms of the given Geometric Progression (a) = 3
The common ratio of the given Geometric Progression (r) = 63 = 2
Let nth term of the given Geometric Progression is 6144.
Then,
⇒ tn = 6144
⇒ a ∙ rn−1 = 6144
⇒ 3 ∙ (2)n−1 = 6144
⇒ (2)n−1 = 2048
⇒ (2)n−1 = 211
⇒ n - 1 = 11
⇒ n = 11 + 1
⇒ n = 12
Therefore, 6144 is the 12th term of the given Geometric Progression.
2. Which term of the Geometric Progression 2, 1, ½, ¼, ............. is 1128?
Solution:
The given Geometric Progression is 2, 1, ½, ¼, .............
The first terms of the given Geometric Progression (a) = 2
The common ratio of the given Geometric Progression (r) = ½
Let nth term of the given Geometric Progression is 1128.
Then,
tn = 1128
⇒ a ∙ rn−1 = 1128
⇒ 2 ∙ (½)n−1 = 1128
⇒ (½)n−1 = (½)7
⇒ n - 2 = 7
⇒ n = 7 + 2
⇒ n = 9
Therefore, 1128 is the 9th term of the given Geometric Progression.
3. Which term of the Geometric Progression 7, 21, 63, 189, 567, ............. is 5103?
Solution:
The given Geometric Progression is 7, 21, 63, 189, 567, .............
The first terms of the given Geometric Progression (a) = 7
The common ratio of the given Geometric Progression (r) = 217 = 3
Let nth term of the given Geometric Progression is 5103.
Then,
tn = 5103
⇒ a ∙ rn−1 = 5103
⇒ 7 ∙ (3)n−1 = 5103
⇒ (3)n−1 = 729
⇒ (3)n−1 = 36
⇒ n - 1 = 6
⇒ n = 6 + 1
⇒ n = 7
Therefore, 5103 is the 7th term of the given Geometric Progression.
● Geometric Progression
11 and 12 Grade Math
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