Geometric Progression: Overview, Questions, Preparation

An ordered collection of items or terms is known as a sequence. A sequence that follows a specific pattern is known as a progression. The numbers that occur in a sequence are terms of the sequence. Geometric progression (GP) is a sequence where the terms follow a pattern. Each succeeding term is the product of its preceding term and a fixed number, the common ratio.

If a sequence g₁, g₂, g₃,...g_n is a GP, then-

g_n+!/g_n= r,

where all the terms are non-zero and n ≥1.

Here, r is constant and is known as the common ratio.

Consider g₁= g; we get the GP:

g, gr, gr², gr³…,

where g is the first term, and r is the common ratio of the GP.

The general term of a GP:

Consider a GP with the first non-zero term ‘g’ and common ratio ‘r’. To obtain the second term, multiply g with r, thus g₂ = gr.

Similarly, to obtain the third term multiply g2 with r. Thus, g3 = g2 r = gr2 , and so on.

Therefore, the pattern suggests that the nth term of a G.P. is given by:

 g_n= g.r^n-1

Thus, G.P. can be written as g, gr, gr²,gr³, …,g.r^n-1

The sum of terms of a GP:

If the first term of a GP is g and the common ratio is r, then Sn is the sum to first n terms of G.P.

S_n= g+ gr+ gr²+gr³ +, …gr^n-1

If we consider r=1, then the above equation becomes:

Sn= g+g+g+g… n times= ng.

If n≠1, we multiply r with the earlier equation
rS_n= gr+ gr²+gr³ +, …grⁿ.

On subtracting this equation from the above equation, we get

(1-r) S_n= g- grⁿ= g(1- rⁿ)

The sum of n terms of a GP is given by:

S_n= {g(1- rⁿ)}/ (1-r)

GP is a part of the chapter ‘Sequence and Series’ in Class 11^th maths. It carries a weightage of around 4-5 marks in the exams.

1. In a G.P., the 3^rd term is 24, and the 6^th term is 192. Find the 10^th term.

Solution.

g₃= 24= gr²

g₆= 192= gr⁵

g₆/ g₃= 192/ 24=> r^3.= 8, so r=2.

Placing the value of r in the first equation, we get

g₃= 24= gr² => g(2)²= 24 => g= 24/4 = 6

Hence, g₁₀= gr⁹= 6 x (2)⁹= 3072.

2.A person creates a pattern with marbles such that the first line has 2 marbles, then 4, then 8, and so on. Find the total number of marbles required to make this pattern in ten lines.

Solution.

Here, first term= g =2 and r=2, n=10

S_n= {g(rⁿ- 1)}/ (r- 1)

S_n= {2 (2¹⁰-1)}/(2-1)

S_n= 2 x (2¹⁰-1) =2046.

3. In the GP 2,8,32, ... up to n terms, which term is 131072?

Solution.

In the given GP, g_n= 131072

First term= a= 2, r= 8/2= 4

g_n = ar^n-1. => 131072= 2 x 4^n-1 => n= 9.

Q: What is the difference between an arithmetic progression and a geometric progression?

Q: What is the common ratio of a GP?

Q: What is the general term of a GP?

Q: What is the sum of m terms of a GP?

Q: What is the geometric mean of two numbers?