[Maths Class Notes] on Geometric Progression Sum of GP Pdf for Exam

In mathematical movement (G.P.), the grouping is mathematical and is a consequence of the amount of G.P. A mathematical series is the amount of the multitude of terms of mathematical arrangement. Prior to going to figure out how to observe the amount of a given Geometric Progression, first know what a GP is exhaustively.

What is Geometric Progression?

In the event that in a succession of terms, each succeeding term is produced by duplicating each first term with a consistent worth, then, at that point, the grouping is known as a mathematical movement. (GP), though the consistent worth is known as the normal proportion. For instance, 2, 4, 8, 16, 32, 64 … is a GP, where the normal proportion is 2.

Amount of Nth terms of G.P.

A mathematical series is an amount of a limitless number of terms with the end goal that the proportion between progressive terms is consistent. In this segment, we will figure out how to track down the amount of mathematical series

Determination of Sum of GP

Since, we know, in a G.P., the normal proportion between the progressive terms is steady, so we will consider a mathematical series of limited terms to infer the equation to track down the amount of Geometric Progression

Let’s assume that there are 10 terms in a sequence. If each term is a multiple of the next term then this sequence is said to be in Geometric Progression or Geometric Sum of G.P. Here, the number which is used to constantly multiply is known as the common ratio. In Arithmetic Progression, when the nth term is subtracted from (n – 1)th term, there will be a constant difference between any two numbers. A.P can be written as:

x, x + b, x + 2b, x + 3b, x + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x + (n – 1) b

Consider the sequence: 

3, 6, 12, 24, 48, . . . . . . . . . . . . . . . . . . . . . and the list goes on..

Observe this:

6/3 = 2

12 / 6 = 2

24 / 12 = 2

48 / 24 = 2

In the same manner: 

Consider the series:  1, ⅓, ⅙, ⅟₁₂,  ⅟₂₄, ⅟₄₈, . . . . . . . . . . . . . . . . . . . . . . . .

⅓ / 1 = ⅟₂

¼ / ½ = ½

⅛ / ¼ = ½

⅟₁₆ / ⅟₈ = ½ 

⅟₂₄ / ⅟₁₆ = ½

⅟₄₈ / ⅟₂₄ = ½

In the examples shown above, you can see that the one thing that is constant is the ratio between any two sequential numbers. These sequences are called Geometric Progression. 

A sequence x₁, x₂, x₃, x₄, . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xₙ is a Geometric Progression, then aₖ₊₁ / aₖ = r  {k is always greater than 1}

Where,

R = a constant ration between two numbers. 

The General Term of Geometric Progression

The nth term of Arithmetic Progression was found out to be:

xₙ  = x + (n – 1) b

In the case of Geometric Progression, let’s assume that x is the first number and “r” is the common ratio between all the numbers.

So, the second term would be: x₂ = x * r 

The third term would be: x₃ = x₂ * r = x * r * r = xr²

In the same manner, the nth term would be xn  = x * rⁿ⁻¹

The general term = xₙ  = x * rⁿ⁻¹

Common Term

Consider the following sequence: x, xr, xr² , xr³, . . . . . . . . . . . . . 

Here, 

The First term is = x

The Second term is = xr

The Third term is = xr₂

Similarly nth term, xₙ = x * rⁿ⁻¹

Hence, the common ratio (r) = (Any given term) / (the next preceding term)

= xₙ  / x * rⁿ⁻¹

= (rⁿ⁻¹) /(rⁿ⁻²)

= r

Thus, the general term of a G.P is given by  rⁿ⁻¹ and the general form of a G.P is x + xr + xr² + . . . . . . . . . . . . . 

Finite and Infinite 

The finite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹

The infinite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹ can be called the finite geometric progression series.

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . can be called as the infinite geometric progression series.

Sum of GP of n terms

Let assume, 

x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹

Let’s assume,

The sum of gp terms =  Sₙ

First Term = x

Ration = r

Now, 

Sₙ = x, xr, xr², xr³, . . . . . . . . . . . . . xrⁿ⁻¹- – – – – – – – – – – – — – – – – – – – – – – (1)

Here, r = 1 or r ≠ 1

If r = 1, then Sₙ = x, x, x, x . . . . . . . . . . . . . . . = nx

If r ≠ 1, then rSₙ = x, xr, xr² , xr³, . . . . . . . . . . . . . xrⁿ⁻¹ + xrⁿ – – – – – – – – – – – – – (2)

Now subtract equation (1) from equation (2), it gives you:

rSₙ- Sₙ = ( xr + xr² + xr³ +. . . . . . . + xrⁿ⁻¹+ xrⁿ) – (x + xr + xr² + xr³+ .. . . . . .+ xrⁿ⁻² + xrⁿ⁻¹)

(r – 1)Sₙ = xrⁿ
– x = x(rⁿ- 1)

Sₙ= x(rⁿ – 1) / (r – 1)

Sₙ= x(1 – rⁿ)/ ( 1 – r)

Sum of n terms in GP

Sₙ= x(1 – rⁿ)/ ( 1 – r) here,r≠1here,r≠1

Solved Problems

Example 1: Find the value of n, if the nth term of Geometric Progression is 2, 4, 8, 16, . . . . . . . . . . . . . . . . , 128.

Solution:

Step 1: Find the ratio

r = 4/2 = 2

Step 2: The first term: x = 2.

xⁿ = xrⁿ⁻¹

128 = 2 x 2ⁿ⁻¹

2ⁿ⁻¹ = 128 / 2

2ⁿ⁻ = 64

2ⁿ⁻¹ = (2)6

n – 1 = 6

n = 6 + 1

n = 7

Hence, the 7th  term here is 128.

Example 2: If the Geometric Progression is 5, 10, 20, 40  . . . . . . . , find the sum of 8 terms.

Solution: 

Given, 

x = 5

n = 8

r = 10/5 = 2

Sum of n terms in gp,

Sₙ = x( r[^{n}]- 1) / (r – 1)

S₈ = 5(2⁸- 1) / (2 – 1)

S₈ = 3 (6561) / 2

S₈ = 3 (256) / 2

S₈ = 768 / 2

S₈ = 384