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Geometric progression or G.P. falls under the category of progressions, which are specific sequences in mathematical terms where each succeeding term is formed by multiplying the corresponding preceding term with a particular fixed number. The word ‘sequence’ depicts a collection of objects in an ordered manner so that all its members can be identified individually. When such a sequence follows a specific pattern in the mathematical application, and that can be determined via calculations, it is called a progression.
Geometric Progression (G.P.): Sequence and Series
Let the components A1, A2, A3,……, An, form a finite sequence with a total of ‘n’ number of terms. Concerning this, the expression A1+A2+A3+……+An is known as the series associated with this particular sequence.
Geometric progression (G.P.)- Definition and General Form
A particular type of sequence is regarded as a geometric progression or G.P. when the ratio of a particular term (except the first one) to its preceding term is the same and remains constant throughout the entire stretch of the sequence. Thus, every term here exhibits a constant ratio to its fellow preceding term in simple words. So, to obtain the next term in a certain G.P., one has to do multiply by the constant ratio while, for finding the previous term, dividing the particular term with the same constant ratio is necessary.
The general form of G.P. is A1, A2, A3,……, An, where ‘A1’ is the first term, ‘A2’ is the second term, ‘A3’ is the third term, and ‘An’ is the nth or general term of this G.P.
An example of a G.P. is 3, 9, 27, 81, 243, …, having a standard ratio of 3.
Geometric Progression (G.P.): Components of G.P.
The essential components of a G.P. are-
- The ‘first term, which is the initial term of a G.P.;
- The ‘common ratio,’ which can be a positive or a negative integer and is the dedicated constant factor that marks the ratio between the successive terms of a G.P., calculated by dividing the two successive terms;
- ‘No. of terms’, generally denoted by ‘n’;
- The ‘general term’ of the G.P., i.e. the nth term.
The nth term can be found by determining the first term of a given G.P. and the standard ratio behind it following the formula,
An=Arn-1
where A is the first term, r is the standard ratio, and n is the total no. of terms.
Geometric Progression (G.P.): Types of G.P.
G.P. is of two basic types based on the number of terms present in the particular progression sequence, namely finite G.P. and infinite G.P.
- A finite G.P. naturally contains a finite number of terms and a defined last term. It can be denoted by the general expression, A, Ar, Ar2, Ar3,……Arn-1.
- An infinite G.P. doesn’t contain any finite number of terms and also lacks a properly defined last term. It can be expressed as A, Ar, Ar2, Ar3,……Arn-1,…….
Geometric Progression (G.P.): How to find the sum of n terms in a G.P.
Let’s suppose A, Ar, Ar2, Ar3,……Arn-1 is a given G.P., where the sum of n terms is denoted by-
Sn = A+Ar+ Ar2 +Ar3+……+ Arn-1
Suppose this is a finite G.P., thus, sum Sn of the first n terms can be calculated by-
Sn = A[(rn-1)/(r-1)] if r ≠ 1
or Sn = nA if r=1,
where A is the first term, r is the common ratio, and n is the total no. of terms in the sequence.
Geometric Progression (G.P.): Geometric mean (G.M.)
The geometric mean or G.M. of any two positive numbers, A and B is denoted by AB, where the corresponding sequence becomes A, G, B; being aligned in the format of a G.P.
Geometric Progression (G.P.): Important formulas related to G.P.
For a particular GP with ‘A’ as the first term, ‘r’ as the common ratio, and ‘n’ number of terms, the following important formulas apply:
- The nth term of the G.P. is xn = Arn-1
- Common ratio, r = xn/ xn-1
- The formula to calculate the sum of the first n terms of the G.P. is:
Sn = A[(rn-1)/(r-1)] if r ≠ 1and r > 1
Sn = A[(1 – rn)/(1 – r)] if r ≠ 1 and r < 1 - The formula to calculate the sum of the terms of an infinite G.P. is:
S∞= A/(1 – r) such that 0 < r < 1
- If A, B, C is a G.P., then B is the geometric mean of A and C; this can be denoted by B2 = A.C or B =√A.C
- Suppose A and r is the first term and the common ratio respectively, belonging to a finite G.P. with n terms. Then, the pth term from the end of the G.P. will be = Arn-p.
Geometric Progression (G.P.): Sample Questions
Question: A geometric progression is given as 2,4,8,16,………., with 1024 being the nth term. Find the value of n.
Sol: Clearly, the common ratio = 2 and the first term, a = 2
We know the nth term,xn=xrn-1
Therefore, 1024 = 2 × 2n-1
Or, 2n-1= 1024/2
Or, 2n-1= 512
Or, 2n-1= 29
Or, n-1 = 9
∴ n = 10 is the required value.
Question: A girl has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Please find the total number of her ancestors during the five preceding generations of her own.
Sol: We know that the ancestors are forming a G.P. in which,
The first term = 2,common ratio = 2 and number of generations i.e., terms = 5
Now, using the sum formula of the n terms of a G.P., i.e.,
Sn = x(rn-1)/(r-1)
We have S5 = 2(25– 1) = 62. Hence 62 is the total number of her ancestors.
Question: Suppose X = the sum, Y = the product, and Z = the sum of reciprocals of all the n terms in a geometric progression, then prove that Y2Zn = Xn.
Sol: Suppose the G.P. is A, Ar, Ar2, Ar3,……Arn-1
So, X = A (rn- 1) / (r-1)
Y = An × r1+2+…+n-1
= An × rn(n-1)/2 [as sum of the first n natural numbers is n(n+1)/2]
= Xn
∴Y2Zn = Xn, hence proved.
Question: Find the 10th and the nth terms of the following geometric progression 5, 25,125,….
Sol: We know that,
The first term, a = 5 and the common ratio, r = 25/5 = 5.
Thus, the 10th term is: a10 = 5× (5)10–1 = 5(5)9 = 510
And the nth term is: an= a×rn–1 = 5×(5)n–1 = 5n.
Question: A man has four friends and he writes a letter to each and then asks everyone to copy the same letter and mail them to four other different people giving instructions that they carry on the chain exactly in a similar fashion. Now, assuming that the chain is not broken anywhere and that it costs 10 paise to mail one single letter, find the total amount of money spent on the postal purpose till all the letters are sent.
Sol: The said letters form a G.P. 4,42,…,48
Therefore, a = 4, r = 4 and n = 8
Now, the sum of the first n terms of a G.P. is-
Sn = a[(rn-1)/(r-1)]
Or, S8 = 4[(48 – 1)/ 4-1]
Or, S8= 4[(65536– 1)/ 3]
Or, S8 = 4 × 21845
∴S8 = 87380
Now, the cost of mailing one letter is 10 paisa.
So, the total cost of mailing 87380 letters=87380 × 10/100 = ? 8738, hence the total money spent on the postal purpose till all the letters are sent.
Question: The 2nd term in a G.P. is 12, and the 4th term is 48. Find out the 10th term.
Sol: Clearly, a2= ar = 12 ... (1)
And, a4 = ar3 = 48 ... (2)
By (2)/(1), we get r = 2
Now, substituting r = 2 in (1), we get a = 6.
Therefore, a10 = 6×(2)9 = 3072.
Question: If the first term of a G.P. is 10 and the common ratio is 3, then find the first three terms of this G.P.
Sol: Given that a=10, r=3
So, the first term, a = 10
The second term= ar= 30 and the third term= ar2=90
Therefore the first three terms of the mentioned G.P. are 10,30, and 90.
Question: If 2,4,8,16,…. is a G.P., then find its 12th term.
Sol: The nth term of GP is given by:
A12 = 2 x 212-1
= 2 x 211 = 4096
Question: Verify whether the given sequence is G.P. or not
64,32,16,…
Sol: Clearly a2/a1=32/64=0.5
Similarly, a3/a2=16/32=0.5, hence verified.
Question: Find the sum of all the terms of this infinite G.P.
1/3, 1/9, 1/27... ∞.
Sol: Given, a = 1/3, |r| = 1/3, thus, |r|< 1
Hence, using the formula:
S∞= a/(1 – r)
Or, S∞ = (1/3)/(1 - 1/3)
Or, S∞ = 1/2, hence the sum of all the terms in the infinite G.P.
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