# Geometric Progression Sum of GP

## Sum of GP

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 * r n-1

The general term = xn = x * r n-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 * r n-1

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

= xn / x * r n-1

= (r n-1) /(r n-2)

= r

Thus, the general term of a G.P is given by r n-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, . . . . . . . . . . . . . xr n-1

The infinite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1,. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1 can be called as the finite geometric progression series.

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

### Sum of GP of n terms

Let assume,

x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-1

Let’s assume,

The sum of gp terms = Sn

First Term = x

Ration = r

Now,

Sn = x, xr, xr² , xr³, . . . . . . . . . . . . . xr n-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, . . . . . . . . . . . . . xr n-1 + xr n – – – – – – – – – – – – – (2)

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

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

(r – 1)Sn = xr n– x = x(r n– 1)

Sn= x(r n– 1) / (r – 1)

Sn= x(1 – r n)/ ( 1 – r)

Sum of n terms in GP

Sn= x(1 – r n)/ ( 1 – r) [here, 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 = xr n-1

128 = 2 x 2 n-1

2 n-1 = 128 / 2

2 n-1 = 64

2 n-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

What is the geometric progression formula?

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

Where,

r = a constant ration between two numbers.

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

x_{n}= 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: x_{2} = x * r

The third term would be: x_{3} = x_{2} * r = x * r * r = xr²

In the same manner, the nth term would be x_{n} = x * r ^{n-1}

The general term = x_{n} = x * r ^{n-1}

How can you define finite and infinite geometric progression?

The finite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xr ^{n-1}

The infinite Geometric Progression terms can be described as:

x, xr, xr² , xr³, . . . . . . . . . . . . . xr ^{n-1},. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x, xr, xr² , xr³, . . . . . . . . . . . . . xr ^{n-1} can be called as the finite geometric progression series.

x, xr, xr² , xr³, . . . . . . . . . . . . . xr ^{n-1} . . . . . . . . . can be called as the infinite geometric progression series.