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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.

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

  • A series in which each preceding term is formed by multiplying it by a constant factor is called a Geometric Progression G. P. The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it.
  • In other words, a sequence, Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC is called a geometric progression
  • If Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC = constant for all n ∈ N.
  • The General form of a G. P. with n terms is a, ar, ar2,…arn -1
    Thus if a = the first term
    r = the common ratio
    Tn = nth term and
    Sn = sum of n terms
  • General term of GP = Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Example.1: Find the 9th term and the general term of the progression.
Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

  • The given sequence is clearly a G. P. with first term a = 1 and common ratio = r = Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC
  • Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSCGeometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC
    Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Question for Geometric Progression: Introduction & Solved Examples
Try yourself:How many terms are there in the GP 5, 20, 80, 320, ..........., 20480? 
View Solution


Common Ratio of G.P.


Consider the sequence a, ar, ar2, ar3,……

First term = a

Second term = ar

Third term = ar2

Similarly, nth term, tn = arn-1

Thus, the common ratio of geometric progression formula is given as:

Common ratio = (Any term) / (Preceding term)

= tn / tn-1

= (arn – 1 ) /(arn – 2)

= r

Thus, the general term of a GP is given by arn-1 and the general form of a GP is a, ar, ar2,…..

For Example: r = t2 / t1 = ar / a = r


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)


(a) Finite Geometric Progression (Finite G.P.)

Suppose a, ar, ar2, ar3,……arn-1 is the given Geometric Progression.

Sum of n terms of a G.P.

The sum of n terms of GP is given by:

Sn = a + ar + ar+ ar+…+ arn-1

The formula to find the sum of n terms of GP is:

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Where,

a is the first term

r is the common ratio

n is the number of terms


(b) Infinite geometric progression (Infinite GP)


Terms of an infinite G.P. can be written as a, ar, ar2, ar3, ……arn-1,…….

a, ar, ar2, ar3, ……arn-1,……. is called infinite geometric series.


Sum of Infinite G.P.


The sum of infinite geometric series is given by:

If a G.P. has infinite terms and -1 < r < 1 or |x| < 1, then sum of infinite G.P. is Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC


Example.2: The inventor of the chess board suggested a reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third and so on, doubling the number of the grains for subsequent squares. How many grains would have to be given to inventor? (There are 64 squares in the chess board).

  • Required number of grains = 1 + 2 + 22 + 23 + ……. To 64 terms =  Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC


Recurring Decimals as Fractions

  • If in the decimal representation a number occurs again and again, then we place a dot (.) on the number and read it as that the number is recurring.
  • Example: 0.5 (read as decimal 5 recurring).
    This mean 0.Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC = 0.55555…….∞
    0. 4 Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC = 0.477777……∞
    These can be converted into fractions as shown in the example given below


Example.3 Find the value in fractions which is same as of Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC


Properties of G.P.

  1. If each term of a GP is multiplied or divided by the same non-zero quantity, then the resulting sequence is also a GP.
    Example: For G.P. is 2, 4, 8, 16, 32…
    Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC
  2. Selection of terms in G.P.
    Sometimes it is required to select a finite number of terms in G.P. It is always convenient if we select the terms in the following manner:Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC
    If the product of the numbers is not given, then the numbers are taken as a, ar, ar2, ar3, ….
  3. Three non-zero numbers a, b, c are in G.P. if and only if
    b2 = ac     or         Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC b is called the geometric mean of a & c
  4. In a GP, the product of terms equidistant from the beginning and end is always same and equal to the product of first and last terms as shown in the next example.
    Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

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, ar2, ar3, and so on. Here, a is the first term and r is the common ratio.
  • The nth term of a GP is Tn = arn-1
  • Common ratio = r = Tn/ Tn-1
  • The formula to calculate the sum of the first n terms of a GP is given by:
    Sn = a[(r– 1)/(r – 1)] if r ≠ 1and r > 1
    Sn = a[(1 – rn)/(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 b2 = 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 = arn-k.

Previous Year Question (2022) related to Geometric Mean is discussed below:


Question: The geometric mean radius of a conductor, having four equal strands with each strand of radius ‘𝑟’, as shown in the figure below, is:

(A) 4 𝑟 

(B) 1.414 𝑟 

(C) 2 𝑟 

(D) 1.723 𝑟  


Solution: GMR is defined as the effective distance over which self magnetic flux linkages occur

For a solid conductor with radius r,

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC

GMR = r' = 0.7788r

GMR is less than the physical radius of the conductor.

In the given figure, standard conductor with four identical strands touching each other is given with equal radius r.

Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC


GMRa1 = (r′×2r×2r×22r)1/4

= 1.722 r

Since each strands are identical.

∴ GMR of conductor will be equal to GMR of strand i.e.,

GMRcond = GMRa1

= 1.722 r

Therefore, correct option is (d)


Solved Examples of Geometric Progression


Q.1: If the first term of a G.P. is 20 and the common ratio is 4. Find the 5th term.
Solution: Given,

First term, a = 20

Common ratio, r = 4

We know, Nth term of G.P.,

an = arn-1

⇒ a5 = 20 × 44

= 20 × 256

= 5120


Q.2: The sum of the first three terms of a G.P. is 21/2 and their product is 27. Find the common ratio.

Solution: Let three terms of G.P. be a/r, a, ar.

Given,

Product of first three terms = 27

⇒ (a/r) (a) (ar) = 27

⇒ a3 = 27

⇒ a = 3.

Sum of first three terms = 21/2

⇒ (a / r + a + ar) = 21/2

⇒ a (1 / r + 1 + 1r) = 21/2

⇒ (1 / r + 1 + 1r) = (21/2)/3 = 7/2

⇒ (r2 + r + 1) = (7/2) r

⇒ r2 – (5/2) r + 1 = 0

⇒ r = 2 and ½


Q.3: Find a Geometric Progress for which the sum of first two terms is -4 and the fifth term is 4 times the third term.

Solution: Let the first term of the geometric series be a and the common ratio is r.

Sum of the first two terms = -4

a + ar = -4 ………………(i)

Fifth term is 4 times the third term.

ar4 = 4ar2

r2 = 4

r = ±2

If we consider r = 2, then putting value of r in eq.(i)

a(1+2) = -4

a = -4/3

ar = -8/3

ar2 = -16/3

Thus, the G.P. is -4/3, -8/3, -16/3, …..


Q.4: The number 2048 is which term in the following Geometric sequence 2, 8, 32, 128, . . . . . . 

Solution: Here a = 2 and r = 4

nth term G.P is an = arn-1

⇒ 2048 = 2 x ( 4)n-1

⇒ 1024 =( 4)n-1

⇒ ( 4)5 = ( 4) n-1

⇒ n = 6


Q.5: In a G.P, the 6th term is 24 and the 13th term is 3/16 then find the 20th term of the sequence.

Solution: Let first term be ‘a’ and common ratio is ‘r’

Given,

a6 = 24 ————- ( i)

a13 = 3/16 ————– ( ii)

⇒ a= a r6-1

a13 = a r13-1

⇒ 24 = a r5

3/16 = a r12

⇒ r7 = 3 / 24 x 16 = 1 / (2)7

⇒ r = 1/2 ———– (iii)

Thus,

⇒ a= 24 = a (1/2)5

⇒ a = 3 x 28

Now a20 = a r20-1

a20 = 3 x 28 x ( 1/2 )19 = 3 / 211


Q.6: Find the sum of the geometric series:

4 – 12 + 36 – 108 + ………….. to 10 terms

Solution: The first term of the given Geometric Progression = a = 4 and its common ratio = r = −12/4 = -3.

Sum of the first 10 terms of geometric series:

S10 = a. (rn – 1/r-1)

= 4. ((-3)10 – 1)/(-3-1)

= – (-3)10 – 1

= – 59048


Q.7: ‘x’ and ‘y’ are two numbers whose AM is 25 and GM is 7. Find the numbers.

Solution: Here x’ and ‘y’ are two numbers then

Arithmetic mean = AM = (x+y)/2

25 = (x+y)/2

x+y = 50 ………(i)

Geometric mean, GM = √(xy)

7 = √(xy

72 = xy

xy = 49 ………..(ii)

Solving equation (i) and (ii), we get;

x = 1 and y = 49.


Q.8: Determine the common ratio r of a geometric progression with the first term is 5 and fourth term is 40.

Solution: Given,

First term, a= 5

Fourth term, a4 = 40

a4/a1 = 40/5

a1r3/a1 = 40/5

r3 = 8

r = 2


Q.9: If the nth term of a GP is 128 and both the first term a and the common ratio 'r' are 2. Find the number of terms in the GP.

Solution: nth term of a GP, an = 128

First term of GP, a = 2

Common ratio, r = 2

Nth term of G.P., an = a.rn-1

128 = 2.2n-1

64 = 2n-1

26 = 2n-1

n- 1 = 6

n = 7

Therefore, there are 7 terms in GP.


Q.10: What is the sum of infinite geometric series with first term equal to 1 and common ratio is ½?

Solution: By the formula of sum of infinite geometric series, we have;

S = a1. 1/(1 – r)

S = 1. 1/(1-½)

S = 1/(½)

S = 2

Hence, the required sum is 2.


Frequently Asked Questions on Geometric Progression

1. What is a Geometric Progression?
Answer: 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.

2. Give an example of Geometric Progression.

Answer: The example of GP is: 3, 6, 12, 24, 48, 96,…


3. What is the general form of GP?

Answer: The general form of a Geometric Progression (GP) is given by a, ar, ar2, ar3, ar4,…,arn-1

a = First term
r = common ratio
arn-1 = nth term


4. What is the common ratio in GP?
Answer: 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


5. What is not a geometric progression?

Answer: If the common ratio between each term of a geometric progression is not equal then it is not a GP.


6. What is the sum of a geometric series?

Answer: If a, ar, ar2, ar3,……arn-1 is the given Geometric Progression, then the formula to find sum of GP is:

Sn = a + ar + ar+ ar+…+ arn-1

Or

Sn= a[(r– 1)/(r – 1)] where r ≠ 1 and r > 1

The document Geometric Progression: Introduction & Solved Examples | CSAT Preparation - UPSC is a part of the UPSC Course CSAT Preparation.
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FAQs on Geometric Progression: Introduction & Solved Examples - CSAT Preparation - UPSC

1. What is the common ratio of a geometric progression?
Ans. The common ratio of a geometric progression (G.P.) is the constant ratio between any two consecutive terms. It is denoted by 'r' and is obtained by dividing any term by its preceding term.
2. What are the types of geometric progressions?
Ans. There are two types of geometric progressions: finite geometric progressions and infinite geometric progressions. In a finite geometric progression, the series has a specific number of terms, while in an infinite geometric progression, the series continues indefinitely.
3. How can recurring decimals be expressed as fractions using geometric progression?
Ans. To express a recurring decimal as a fraction using geometric progression, we first convert the recurring decimal into an equation. We then solve the equation to obtain a value, which is the sum of an infinite geometric progression. Finally, we use the formula for the sum of an infinite geometric progression to simplify the expression and represent it as a fraction.
4. What are the properties of a geometric progression?
Ans. The properties of a geometric progression include: - Each term is obtained by multiplying the preceding term by a constant ratio. - The sum of the first 'n' terms of a geometric progression can be calculated using the formula: Sn = a(1 - r^n) / (1 - r), where 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms. - If the absolute value of the common ratio is less than 1, the geometric progression converges to a finite limit. If the absolute value is greater than or equal to 1, the progression diverges. - The product of any number of terms in a geometric progression is equal to the product of the first term and the common ratio raised to the power of the number of terms.
5. What are some common formulas used in geometric progressions?
Ans. Some common formulas used in geometric progressions include: - nth term of a geometric progression: an = a * r^(n-1), where 'an' is the nth term, 'a' is the first term, 'r' is the common ratio, and 'n' is the term number. - Sum of the first 'n' terms of a geometric progression: Sn = a(1 - r^n) / (1 - r), where 'Sn' is the sum, 'a' is the first term, 'r' is the common ratio, and 'n' is the number of terms. - Sum of an infinite geometric progression: S = a / (1 - r), where 'S' is the sum, 'a' is the first term, and 'r' is the common ratio.
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