In Maths, Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern. Also, learn arithmetic progression here. The common ratio multiplied here to each term to get the next term is a non-zero number. An example of a Geometric sequence is 2, 4, 8, 16, 32, 64, …, where the common ratio is 2.
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What is Geometric Sequence?
A geometric progression or a geometric sequence is the sequence, in which each term is varied by another by a common ratio. The next term of the sequence is produced when we multiply a constant (which is non-zero) to the preceding term. It is represented by:
a, ar, ar^{2}, ar^{3}, ar^{4}, and so on.
Where a is the first term and r is the common ratio.
Note: It is to be noted that when we divide any succeeding term from its preceding term, then we get the value equal to the common ratio.
Suppose we divide the 3rd term by the 2nd term we get:
ar^{2}/ar = r
In the same way:
ar^{3}/ar^{2} = r
Properties of Geometric Progression (GP)
Some of the important properties of GP are listed below:
- Three non-zero terms a, b, c are in GP if and only if b^{2} = ac
- In a GP,
Three consecutive terms can be taken as a/r, a, ar
Four consecutive terms can be taken as a/r^{3}, a/r, ar, ar^{3}
Five consecutive terms can be taken as a/r^{2}, a/r, a, ar, ar^{2} - In a finite GP, the product of the terms equidistant from the beginning and the end is the same
That means, t_{1}.t_{n} = t_{2}.t_{n-1} = t_{3}.t_{n-2} = ….. - If each term of a GP is multiplied or divided by a non-zero constant, then the resulting sequence is also a GP with the same common ratio
- The product and quotient of two GP’s is again a GP
- If each term of a GP is raised to the power by the same non-zero quantity, the resultant sequence is also a GP
- If a_{1}, a_{2}, a_{3},… is a GP of positive terms then log a_{1}, log a_{2}, log a_{3},… is an AP (arithmetic progression) and vice versa
General Form of Geometric Progression
The general form of Geometric Progression is:
a, ar, ar^{2}, ar^{3}, ar^{4},…, ar^{n-1}
Where,
a = First term
r = common ratio
ar^{n-1} = nth term
General Term or Nth Term of Geometric Progression
Let a be the first term and r be the common ratio for a Geometric Sequence.
Then, the second term, a_{2} = a × r = ar
Third term, a_{3} = a_{2} × r = ar × r = ar^{2}
Similarly, nth term, a_{n} = ar^{n-1}
Therefore, the formula to find the nth term of GP is:
a_{n} = t_{n} = ar^{n-1} |
Note: The nth term is the last term of finite GP.
Common Ratio of GP
Consider the sequence a, ar, ar^{2}, ar^{3},……
First term = a
Second term = ar
Third term = ar^{2}
Similarly, nth term, t_{n} = ar^{n-1}
Thus, the common ratio of geometric progression formula is given as:
Common ratio = (Any term) / (Preceding term)
= t_{n} / t_{n-1}
= (ar^{n – 1} ) /(ar^{n – 2})
= r
Thus, the general term of a GP is given by ar^{n-1} and the general form of a GP is a, ar, ar^{2},…..
For Example: r = t_{2} / t_{1} = ar / a = r
Sum of N term of GP
Suppose a, ar, ar^{2}, ar^{3},……ar^{n-1} is the given Geometric Progression.
Then the sum of n terms of GP is given by:
S_{n} = a + ar + ar^{2 }+ ar^{3 }+…+ ar^{n-1}
The formula to find the sum of n terms of GP is:
S_{n} = a[(r^{n }– 1)/(r – 1)] if r ≠ 1 and r > 1 |
Where
a is the first term
r is the common ratio
n is the number of terms
Also, if the common ratio is equal to 1, then the sum of the GP is given by:
S_{n} = na if r = 1 |
Types of Geometric Progression
Geometric progression can be divided into two types based on the number of terms it has. They are:
- Finite geometric progression (Finite GP)
- Infinite geometric progression (Infinite GP)
These two GPs are explained below with their representations and the formulas to find the sum.
Finite Geometric Progression
The terms of a finite G.P. can be written as a, ar, ar^{2}, ar^{3},……ar^{n-1}
a, ar, ar^{2}, ar^{3},……ar^{n-1} is called finite geometric series.
The sum of finite Geometric series is given by:
S_{n} = a[(r^{n }– 1)/(r – 1)] if r ≠ 1 and r > 1 |
Infinite Geometric Progression
Terms of an infinite G.P. can be written as a, ar, ar^{2}, ar^{3}, ……ar^{n-1},…….
a, ar, ar^{2}, ar^{3}, ……ar^{n-1},……. is called infinite geometric series.
The sum of infinite geometric series is given by:
Geometric Progression Formulas
The list of formulas related to GP is given below which will help in solving different types of problems.
- The general form of terms of a GP is a, ar, ar^{2}, ar^{3}, and so on. Here, a is the first term and r is the common ratio.
- The nth term of a GP is T_{n} = ar^{n-1}
- Common ratio = r = T_{n}/ T_{n-1}
- The formula to calculate the sum of the first n terms of a GP is given by:
S_{n} = a[(r^{n }– 1)/(r – 1)] if r ≠ 1and r > 1
S_{n} = a[(1 – r^{n})/(1 – r)] if r ≠ 1 and r < 1 - The nth term from the end of the GP with the last term l and common ratio r = l/ [r(n – 1)].
- The sum of infinite, i.e. the sum of a GP with infinite terms is S_{∞}= a/(1 – r) such that 0 < r < 1.
- If three quantities are in GP, then the middle one is called the geometric mean of the other two terms.
- If a, b and c are three quantities in GP, then and b is the geometric mean of a and c. This can be written as b^{2} = ac or b =√ac
- Suppose a and r be the first term and common ratio respectively of a finite GP with n terms. Thus, the kth term from the end of the GP will be = ar^{n-k}.
Video Lesson
Applied Concept – Sum of Infinite Terms of G.P.
Solved Examples of Geometric Progression
Question 1: If the first term is 10 and the common ratio of a GP is 3, then write the first five terms of GP.
Solution: Given,
First term, a = 10
Common ratio, r = 3
We know the general form of GP for first five terms is given by:
a, ar, ar^{2}, ar^{3}, ar^{4}
a = 10
ar = 10 × 3 = 30
ar^{2} = 10 × 3^{2} = 10 × 9 = 90
ar^{3} = 10 × 3^{3} = 270
ar^{4} = 10 × 3^{4} = 810
Therefore, the first five terms of GP with 10 as the first term and 3 as the common ratio are:
10, 30, 90, 270 and 810
Question 2: Find the sum of GP: 10, 30, 90, 270 and 810, using formula.
Solution: Given GP is 10, 30, 90, 270 and 810
First term, a = 10
Common ratio, r = 30/10 = 3 > 1
Number of terms, n = 5
Sum of GP is given by;
S_{n} = a[(r^{n }– 1)/(r – 1)]
S_{5} = 10[(3^{5 }– 1)/(3 – 1)]
= 10[(243 – 1)/2]
= 10[242/2]
= 10 × 121
= 1210
Check: 10 + 30 + 90 + 270 + 810 = 1210
Question 3: If 2, 4, 8,…., is the GP, then find its 10th term.
Solution: The nth term of GP is given by:
2, 4, 8,….
Here, a = 2 and r = 4/2 = 2
a_{n} = ar^{n-1}
Therefore,
a_{10} = 2 x 2^{10 – 1}
= 2 × 2^{9}
Practice Problems on Geometric Progression
- Find the equivalent fraction of the recurring decimal 0.595959…..
- What is the 12th term of the sequence 4, -8, 16, -64,….?
- Check whether the given sequence is GP.
27, 9, 3, … - Write the first five terms of a GP whose first term is 3 and the common ratio is 2.
Frequently Asked Questions on Geometric Progression
What is a Geometric Progression?
Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio.
Give an example of Geometric Progression.
The example of GP is: 3, 6, 12, 24, 48, 96,…
What is the general form of GP?
The general form of a Geometric Progression (GP) is given by a, ar, ar^{2}, ar^{3}, ar^{4},…,ar^{n-1}
a = First term
r = common ratio
ar^{n-1} = nth term
What is the common ratio in GP?
The common multiple between each successive term and preceding term in a GP is the common ratio. It is a constant value that is multiplied by each term to get the next term in the Geometric series. If a is the first term and ar is the next term, then the common ratio is equal to:
ar/a = r
What is not a geometric progression?
If the common ratio between each term of a geometric progression is not equal then it is not a GP.
What is the sum of a geometric series?
If a, ar, ar^{2}, ar^{3},……ar^{n-1} is the given Geometric Progression, then the formula to find sum of GP is:
S_{n} = a + ar + ar^{2 }+ ar^{3 }+…+ ar^{n-1}
Or
S_{n}= a[(r^{n }– 1)/(r – 1)] where r ≠ 1 and r > 1