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.
Example.1: Find the 9th term and the general term of the progression.
 The given sequence is clearly a G. P. with first term a = 1 and common ratio = r =
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^{n1}
Thus, the common ratio of geometric progression formula is given as:
Common ratio = (Any term) / (Preceding term)
= t_{n} / t_{n1}
= (ar^{n – 1} ) /(ar^{n – 2})
= r
Thus, the general term of a GP is given by ar^{n1} and the general form of a GP is a, ar, ar^{2},…..
For Example: r = t_{2} / t_{1} = ar / a = r
Geometric progression can be divided into two types based on the number of terms it has. They are:
Suppose a, ar, ar^{2}, ar^{3},……ar^{n1} is the given Geometric Progression.
The sum of n terms of GP is given by:
S_{n} = a + ar + ar^{2 }+ ar^{3 }+…+ ar^{n1}
The formula to find the sum of n terms of GP is:
Where,
a is the first term
r is the common ratio
n is the number of terms
Terms of an infinite G.P. can be written as a, ar, ar^{2}, ar^{3}, ……ar^{n1},…….
a, ar, ar^{2}, ar^{3}, ……ar^{n1},……. is called infinite geometric series.
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
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 + 2^{2} + 2^{3} + ……. To 64 terms =
Example.3 Find the value in fractions which is same as of
The list of formulas related to GP is given below which will help in solving different types of problems.
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,
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.
GMR_{a1 }= (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.,
GMR_{cond} = GMR_{a1}
= 1.722 r
Therefore, correct option is (d)
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.,
a_{n} = ar_{n1}
⇒ a_{5} = 20 × 4_{4}
= 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
⇒ a^{3} = 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
⇒ (r^{2} + r + 1) = (7/2) r
⇒ r^{2} – (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.
ar^{4} = 4ar^{2}
r^{2} = 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
ar^{2} = 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 = ar^{n1}
⇒ 2048 = 2 x ( 4)^{n1}
⇒ 1024 =( 4)^{n1}
⇒ ( 4)^{5} = ( 4) ^{n1}
⇒ 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,
a_{6} = 24 ———— ( i)
a_{13} = 3/16 ————– ( ii)
⇒ a_{6 }= a r^{61}
a_{13} = a r^{131}
⇒ 24 = a r_{5}
3/16 = a r^{12}
⇒ r^{7}_{ }= 3 / 24 x 16 = 1 / (2)^{7}
⇒ r = 1/2 ———– (iii)
Thus,
⇒ a_{6 }= 24 = a (1/2)^{5}
⇒ a = 3 x 2^{8}
Now a_{20 }= a r^{201}
a_{20 }= 3 x 2^{8} x ( 1/2 )^{19} = 3 / 2^{11}
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:
S_{10 }= a. (r^{n} – 1/r1)
= 4. ((3)^{10} – 1)/(31)
= – (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
7^{2} = 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_{1 }= 5
Fourth term, a_{4} = 40
a_{4}/a_{1} = 40/5
a_{1}r^{3}/a_{1} = 40/5
r^{3} = 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, a_{n} = 128
First term of GP, a = 2
Common ratio, r = 2
Nth term of G.P., an = a.r^{n1}
128 = 2.2^{n1}
64 = 2^{n1}
2^{6} = 2^{n1}
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 = a_{1}._{ }1/(1 – r)
S = 1. 1/(1½)
S = 1/(½)
S = 2
Hence, the required sum is 2.
Answer: The example of GP is: 3, 6, 12, 24, 48, 96,…
Answer: The general form of a Geometric Progression (GP) is given by a, ar, ar^{2}, ar^{3}, ar^{4},…,ar^{n1}
a = First term
r = common ratio
ar^{n1} = nth term
ar/a = r
Answer: If the common ratio between each term of a geometric progression is not equal then it is not a GP.
Answer: If a, ar, ar^{2}, ar^{3},……ar^{n1} is the given Geometric Progression, then the formula to find sum of GP is:
S_{n} = a + ar + ar^{2 }+ ar^{3 }+…+ ar^{n1}
Or
S_{n}= a[(r^{n }– 1)/(r – 1)] where r ≠ 1 and r > 1
197 videos151 docs200 tests

1. What is the common ratio of a geometric progression? 
2. What are the types of geometric progressions? 
3. How can recurring decimals be expressed as fractions using geometric progression? 
4. What are the properties of a geometric progression? 
5. What are some common formulas used in geometric progressions? 
197 videos151 docs200 tests


Explore Courses for UPSC exam
