nth term of a Geometric Progression

n^th 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……ar^n-1

Where a = the first term

r = common ratio

Now let us learn how to find the n^th term in a geometric progression.

Formula for the n^th 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 n^th 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 n^th term of a GP is T_n = ar^n-1

Solved Examples:

1. If 5, 20, 80, 320… is a geometric progression, find the 7^th 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 n^th term of a geometric progression is T_n = ar^n-1

T₇  =5 ✕ 4⁶

T₇  =5 ✕ 4,096

T  =20,480

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

Solution:

            We know that the n^th term of a geometric progression can be expressed as T_n = ar^n-1

So, T₃ = 40 =  ar^3-1 and

T₅ = 160 = ar^5-1

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 8^th term of the GP, we can use the formula  T_n = ar^n-1

T₈ = 10 x 2^8-1

`T₈ = 10 x 128

T₈ = 1280.

3. 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 n^th term of a GP is T_n = ar^n-1

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

64= 2^n-1 = 2⁶

n = 6+1 = 7.

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