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Geometric progression is a series in which each consecutive element is derived by multiplying the preceding element by a constant known as the common ratio. Generally, the geometric progression series is represented in the manner of a, ar, ar2.... where a is the first term and r is the common ratio of the sequence.
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Keyterms: Series, ratio, Sequence, Finite geometric progression, Infinite geometric progression, Common ratio
Also Read: Pascal’s Triangle
What is Geometric Progression?
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Geometric progression is a series in which each consecutive element is derived by multiplying the preceding element by a constant known as the common ratio, which is indicated by the letter r. It can be negative or positive.
Geometric Progression
The term "series" refers to the arranging of items in such a way that they can all be identified separately. When this series follows a particular pattern in the mathematical application which can be determined by calculation, it is known as progression.
Generally, the geometric progression series is represented as,
a, ar, ar2...
Where,
a → First term
r → Common ratio of the sequence.
For example: In the given series- 1,2,4,8,16,32 and 64.
The common ratio or r is 2. Each number is multiplied by 2 here or we can say the succeeding number is twice the preceding number.
Geometric Progression Common Ratio
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Types of geometric progression
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There are two types of geometric progression:
- Finite geometric progression
- Infinite geometric progression
Finite Geometric Progression
It is a series that consists of finite numbers of terms which means the last term of the series is defined. It can be denoted by the general expression,
A, Ar, Ar2, Ar3,……Arn-1.
For example 1/2,1/4,1/8,1/16,...,1/32768 is a finite geometric series where the last term is 1/32768.
Infinite Geometric Progression
Infinite geometric progression series is the opposite of finite geometric series. It is a series that consists of infinite numbers of terms which means the last term of the series is not defined. It can be denoted by the general expression as
A, Ar, Ar2, Ar3,……Arn-1,…….
For example, 3, −6, +12, −24, +... is an infinite series where the last term is not defined.
Formula of geometric progression
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The formula for geometric progression is used to find the nth term in the sequence. For this, we need a common ratio or ‘r’, and the first term ‘a’. The formula is
an = arn-1
Where,
a → First term
r → Common ratio
n → Number of terms.
Read More: Sequence and Series
Sum Formula for Geometric Progression
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The sum formula is used to find the sum of all the members in the given series. The sum formula for finite geometric progression and infinite geometric progression is different.
For finite geometric progression, the sum formula is -
Sn = a\(\frac {(1-r^n)}{(1-r)}\) for r≠1
Where,
n → Number of Terms
a → 1st Term
r → Common Ratio
For infinite geometric progression, there are two formulas based on two different situations.
Case 1: When the value of r is less than one
The sum formula is given as follows.
S∞ = \(\frac {a} {(1-r)}\)
Case 1: When the value of r is more than one
The series does not converge and has no sum in this situation.
Properties of Geometric Progression
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- In GP when each term is divided or multiplied by a non-zero quantity, the new sequence is also in G.P with the same common difference.
- When all the terms are reversed in GP, a new GP sequence is formed.
- When all the terms in GP are increased with the same power the new series is also GP.
- The three non-zero terms x, y, and z are in G.P. if y2 = xz
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Things to remember
- A geometric progression is a series in which each consecutive element is derived by multiplying the preceding element by a constant known as the common ratio, which is indicated by the letter r. It can be negative or positive.
- The geometric progression series is represented in the manner of a, ar, ar2... where a is the first term and r is the common ratio of the sequence.
- There are two types of GP namely Infinite progression and Finite progression.
- Finite geometric progression is a series that consists of finite numbers of terms which means the last term of the series is defined.
- Infinite geometric progression series is the opposite of finite geometric series. It is a series that consists of infinite numbers of terms which means the last term of the series is not defined.
Read More: NCERT Solutions for Class 11 Math Sequences and Series
Previous Year Questions
- If the 2nd, 5th and 9th terms of a non-constant… (JEE Main 2016)
- If 4x = 16y = 64z, then… (JKCET 2011)
- Sum of n terms of the series 1 + 11…
- Sum to 10 terms of the series 1+2…
- The 10th common term between the series
- The sum of squares of deviations for 10 observations taken from mean 50…
- If Sm denotes the sum of first m terms of a G.P. of n terms
- The 100th term of the sequence 1, 2, 2, 3, 3, 3,....
- A sequence containing a finite number of terms is called a…
- For a positive integer n let a(n)... (JEE Advanced 1999)
Sample questions
Ques. How many terms in the 1 + 3+ 9+.... sequence add up to 121? (2 marks)
Ans. The sum of n GP terms is a * (rn – 1)/(r – 1)
In this case, a = 1 and r = 3.
We get n = 5 by substituting values in the equation.
Ques. What is the main difference between geometric progression and arithmetic progression? (2 marks)
Ans. A sequence is a group of integers that are arranged in a specified order. The difference between two consecutive terms in an arithmetic sequence is always the same. A geometric sequence, on the other hand, is one in which the ratio of two consecutive words is constant.
Ques. Every hour, the number of germs in a particular culture doubles. Initially, the culture had three microorganisms. At the end of the sixth hour, what would be the total number of bacteria? (2 marks)
Ans. The number of bacteria, in this case, follows a geometric development with the first term equaling 3 and the common ratio r equaling 2.
So, at the conclusion of the sixth hour, the total number of bacteria will be the sum of the first six terms of this progression given by S6
S6 = 3(26−1)/(2−1)
→ 3(64−1)/1
→ 3×63
→ 189
Ques. What is the formula for calculating the sum in a geometric progression that is infinite? (2 marks)
Ans. The formula S=a1/(1r) is used to compute the sum of an infinite geometric series including ratios with absolute values smaller than one. The first term is a1, and the common ratio is r.
Ques. Calculate the total of the terms in this infinite geometric progression as follows: 1/3, 1/9, 1/27, and so on… (2 marks)
Ans. Here,
a = 1/3, |r| = 1/3 and |r|< 1
As a result, the sum of an infinite geometric progression is calculated as follows:
S = a/(1-r)
S = (1/3)(1 - 1/3)
S = 1/2
Ques. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle. (2 marks)
Ans. Let the given equilateral triangle be Δ ABC with each side of 20 cm.
By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of the side of ΔABC.
Continuing in this way, we get a set of equilateral triangles with sides equal to half of the side of the previous triangle.
Now,
Perimeter of first triangle = 20 x 3 = 60 cm;
Perimeter of second triangle = 10 x 3 = 30 cm;
Perimeter of third triangle = 5×3 = 15 cm;
Clearly 60, 30,15 form a GP with a=60 and r=30/60 =½
We have to find the perimeter of the sixth inscribed triangle, that is we have to find the sixth term of the GP.
Perimeter of 6th inscribed triangle
a6 = ar 6-1 = 60 x (½)55 = 60/32 = 15/8 cm
Ques. In a potato race, 20 potatoes are placed in a line at intervals of 4 m with the first potato 24 m from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes? (2 marks)
Ans. Distance travelled to bring first potato = 24 + 24 = 2 x 24 = 48 m
Distance travelled to bring second potato = 2(24 + 4) = 2 x 28 = 56 m
Distance travelled to bring the third potato = 2(24 + 4 + 4) = 2 X 32 = 64 m; and so on…
Clearly, 48, 56, 64,… is an A.P. with first term 48 and common difference 8. Also, the number of terms is 20.
Total distance run in bringing back all the potatoes,
S20 = 20/2 [ 2 x48 + (20-1) x 8 ] = 20[ 48 + 76] = 20 x 124 = 2480m
Ques. Find the sum of the series (33 – 23) + (53 – 43) + (73 – 63) + … to (3 marks)
(i) n terms
(ii) 10 terms
Ans. Given series is: (33 — 23) + (53 – 43) + (73 – 63) + … n terms
Tn = (2n + 1)3 – (2n)3 = (2n + 1 – 2n)[(2n + 1)2 + (2n + 1)2n +(2n)2]
= 12n2 + 6n + 1
(i) Sum of n terms,
Sn = \(\sum_{n=1}^n\)(12n2 + 6n + 1) = 12 . \(\frac{n(n +1)(2n+1)}{6} + \frac{6n(n+1)}{2} + n\)
= 2n(n + 1)(2n + 1) + 3n(n + 1) + n
= 2n(2n2 + 3n + 1) + 3n2 + 3n + n
= 4n3 + 9n2 + 6n
(ii) Sum of 10 terms, S10 = 4 x (10)3 + 9 x (10)2 + 6 x 10
= 4000 + 900 + 60 = 4960
Ques. The 3rd and the 8th term of a G. P. are 4 and 128 respectively. Find the G. P. (2 marks)
Ans. Let a be the first term and r be the common ratio of the G. P., then ar3 = 8 …(1)
ar7=128 …(2)
Dividing 2 by 1:
r4 = 16
r = 2
putting r=2 in equation(1)
we get a=1.
So the required GP= 1,2,4,8,16…
Ques. Which term of the G. P.: 6, –12, 24, – 48, ... is 384? (2 marks)
Ans. Let the term be nth.
(ar)n-1 = 384
a=6, and r= (-2)
Solving for n:
6 (-2)n-1 = 384
(-2)n-1= 64
n-1 = 6
n=7
Ques. Find the sum of GP.: 1, 2,4,8, ... up to the 10th term. (2 marks)
Ans. Using the formula for summation of n terms of a GP:
Sn = a(rn-1)/(r-1)
Putting the value of a=1, r =2 , n=10
We get the solution :
210 - 1
Ques. Find the sum of an infinite GP 3,1,1/3,….? (2 marks)
Ans. The sum of an infinite GP is given by,
Sn = a / 1 - r
a = 3, r=1/3.
Putting the values in the equation :
3 / (1-1/3)= 9/2
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