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nth term of a Geometric Progression

nth term of a Geometric Progression

The series in which the ratio between any two consecutive numbers is the same is known as geometric progression. For example, consider a series containing the terms 1, 2, 4, 8, 16, and so on. The ratio of any two consecutive numbers i.e \frac{2}{1} = \frac{4}{2} = \frac{16}{8} = 2

So, a geometric progression can be expressed in the form of a, ar, ar², ar³, ar4……arn-1

Where a = the first term

r = common ratio

Now let us learn how to find the nth term in a geometric progression.

Formula for the nth term of a Geometric Progression

Consider a geometric progression having the terms 5, 15, 45, 135, 405, and 1215.

The first term is 5 and the common ratio of the above series is 3.

 

Now, we can write the different terms in the above series as below.

T1 = 5  = a

T2 = 15 = 5X3 = ar

T3 = 45 = 5X32 = ar2

                T4 = 135 = 5X33 = ar3

 

Can you spot the pattern here? The pattern that is emerging is that the nth term in a geometric progression is the product of the first term and the common ratio raised to (n-1).

 

So, the formula to calculate the nth term of a GP is Tn = arn-1

 

Solved Examples:

  1. If 5, 20, 80, 320… is a geometric progression, find the 7th term of the series.

 

Solution:

From the given question, we know that the first term a = 5 and the common ratio r =20/5=4

The formula to calculate the nth term of a geometric progression is Tn = arn-1

T7  =5 ✕ 46

T7  =5 ✕ 4,096

T  =20,480

  1. If the 3rd term and 5th term of a GP are 40 and 160, find the 8th term of the GP.

 

Solution:

            We know that the nth term of a geometric progression can be expressed as Tn = arn-1

 

So, T3 = 40 =  ar3-1 and

T5 = 160 = ar5-1

40 = ar2                                                                                                                           (1)

160 = ar4                                                                                                                     (2)

Now, dividing (2) by (1), we get

4 = r2

r = √4

r = 2.

Substituting r = 2 in (1), we get

40 = ax22

a = 10.

To find the 8th term of the GP, we can use the formula  Tn = arn-1

T8 = 10 x 28-1

`T8 = 10 x 128

T8 = 1280.

 

  1. How many terms are there in a GP containing the terms 3, 6, 12, ….. 192?

 

Solution:

In the given question, a= 3 and r =6/3=2.  and the last term is 192.

We know the nth term of a GP is Tn = arn-1

In the given question, Tn = 192 = 3 x 2n-1

64= 2n-1 = 26

n = 6+1 = 7.

Hence, there are 7 terms in the given GP series.

 

Ready to get started ?

Frequently Asked Questions 

1. How to calculate the sum of a GP series?

Ans We can calculate the sum of a geometric progression series using the following formulas:

S = (a (1- r^n  ))/(1-r) ( where r <1)

When r >1, S = (a ( r^n-1 ))/(r-1)

 

  1. How to calculate the sum of an infinite GP series?

Ans The sum of an infinite GP series is calculated using the formula S = (a (1- r^n  ))/(1-r)  where |r| <1 .

 

  1. How to find the value of r in a GP?

Ans: r is the common ratio of any two consecutive numbers in a GP series. The value of r is constant in any given geometric progression. The value of r can be calculated using the formula Tn/(Tn-1). We can also obtain the value of r by dividing any term in the GP by its preceding term.

 

  1. How to calculate the number of terms in a GP?

Ans: Where the first term, common ratio, and the last term of a geometric progression are given, we can find the number of terms in the given GP by substituting the given values in the formula  Tn = arn-1 and solving for n.