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Position of a Term in a Geometric Progression

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 = arn1.

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 rn1 = 6144

⇒ 3 (2)n1 = 6144

⇒ (2)n1 = 2048

⇒ (2)n1 = 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 rn1 = 1128

⇒ 2 (½)n1 = 1128

⇒ (½)n1 = (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 rn1 = 5103

⇒ 7 (3)n1 = 5103

⇒ (3)n1 = 729

⇒ (3)n1 = 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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