Mathematics SEQUENCES AND SERIES - Geometric Progression (G . P.), General term ,Sum to n terms of a G .P. and Relationship Between A.M. and G.M.
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### Topics Covered

star Geometric Progression (G . P.)
star General term of a G .P.
star Sum to n terms of a G .P.
star Relationship Between A.M. and G.M.

### Geometric Progression (G . P.)

\color{fuchsia} \(★ mathbf\ul"Geometric Progression (G . P.)")

A sequence color(blue)(a_1, a_2, a_3, …, a_n, … ) is called geometric progression,

if each term is non-zero and color(red)(a_(k+1)/a_k = r) (constant ratio ), for color(red)(k ≥ 1.)

\color{red} ✍️ By letting color(blue)(a_1 = a), we obtain a geometric progression, color(red)(a, ar, ar^2, ar^3,…...,)

where a is called the color(blue)"first term" and r is called the color(blue)"common ratio" of the G.P.

We shall use the following notations with these formulae:

color(red)(a) =  the first term,  \ \ color(red)(r) = the common ratio,  \ \ color(red)(l) = the last term,

color(red)(n) = the numbers of terms,

color(red)(S_n) = the sum of n terms.

color(red)("Let us consider the following sequences:")

(i) color(blue)(2,4,8,16,.............,) (ii) color(blue)(1/9 , -1/ 27, 1/81 , -1/243 ..............) (iii) color(blue)(.01,.0001,.000001,..................)

In (i) we have a_1=2 , a_2/a_1=2, a_3/a_2=2, a_4/a_3=2 and so on. In (ii), we observe, a_1=1/9 , a_2/a_1=-1/3, a_3/a_2=-1/3 , a_4/a_3=-1/3 and so on.

Similarly, state how do the terms in (iii) progress

In (i), this constant ratio is 2; in (ii) it is -1/3 and in (iii) the constant ratio is 0.01.

Common ratio in geometric progression (i), (ii) and (iii) above are 2, -1/3 and 0.01, respectively.

Such sequences are called geometric sequence or geometric progression abbreviated as G.P

### General term of a G .P.

Let us consider a G.P. with first non-zero term ‘a’ and common ratio ‘r’.

The second term is obtained by multiplying a by r, thus color(blue)(a_2 = ar).

Similarly, third term is obtained by multiplying a_2 by r. Thus, color(blue)(a_3 = a_2r = ar^2), and so on.

We write below these and few more terms.
1st term color(blue)(= a_1 = a = ar^(1–1)),

2nd term color(blue)(= a_2 = ar = ar^(2–1)),

3rd term color(blue)(= a_3 = ar^2 = ar^(3–1)),

4th term color(blue)(= a_4 = ar^3 = ar^(4–1)),

5th term color(blue)(= a_5 = ar^4 = ar^(5–1)),

Do you see a pattern? What will be 16^(th) term?

color(blue)(a_16 = ar^(16–1) = ar^15)

Therefore, the pattern suggests that the n^(th) term of a G.P. is given by color(red)(a _n =ar^( n−1).)

Thus, a, G.P. can be written as color(blue)(a, ar, ar^2, ar^3, … ar^(n – 1)) \ \ \ \ \; color(red)(a, ar, ar^2,...,ar^(n – 1)....... ;) according as G.P. is finite or infinite, respectively.

The series color(blue)(a + ar + ar^2 + ... + ar^(n–1)) or color(red)(a + ar + ar^2 + ... + ar^(n–1) +....... ) are called finite or infinite geometric series, respectively.

### Sum to n terms of a G .P.

Let the first term of a G.P. be a and the common ratio be r.

Let us denote by S_n the sum to first n terms of G.P. Then

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(S_n = a + ar + ar^2 +...+ ar^(n–1)) .................... (1)

color(red)(Case 1) If color(blue)(r = 1), we have color(blue)(S_n = a + a + a + ... + a) (n terms) color(red)(= na)

color(red)(Case 2) If color(blue)(r ≠ 1), multiplying (1) by r, we have

 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ color(blue)(rS_n = ar + ar^2 + ar^3 + ... + ar^n) ........................ (2)

Subtracting (2) from (1), we get (1 – r) S_n = a – ar^n = a(1 – r^n)

color(red)(S_n= (a(1 – r^n))/(1-r)) or color(blue)(S_n= (a(r^n-1))/(r-1))
Q 3059812714

Find the 10th and nth terms of the G.P. 5, 25,125,… .

Solution:

Here a = 5 and r = 5. Thus, a_(10) = 5(5)^(10–1) = 5(5)^9 = 5^(10)
and a_n = ar^(n–1) = 5(5)^(n–1) = 5^n
Q 3029012811

Which term of the G.P., 2,8,32, ... up to n terms is 131072?

Solution:

Let 131072 be the nth term of the given G.P. Here a = 2 and r = 4.
Therefore 131072 = a_n = 2(4)^(n – 1) or 65536 = 4^(n – 1)
This gives 48 = 4^(n – 1).
So that n – 1 = 8, i.e., n = 9. Hence, 131072 is the 9th term of the G.P.
Q 3069012815

In a G.P., the 3rd term is 24 and the 6th term is 192.Find the 10th term.

Solution:

Here,a_3 = ar^2 = 24 ..............(1)

and a_6 = ar^5 = 192 .................(2)

Dividing (2) by (1), we get r = 2. Substituting r = 2 in (1), we get a = 6

Hence a_(10) = 6 (2)^9 = 3072
Q 3069112915

Find the sum of first n terms and the sum of first 5 terms of the geometric series 1+2/3+4/9 +.........

Solution:

Here a = 1 and r = 2/3 Therefore S_n = (a (1-r^n))/(1-r) = ([1-2/3])/(1-2/3) = 3 [1-(2/3)^n]

In particular S_5 = 3 [ 1-(2/3)^5] = 3 xx (211)/(243) = (211)/(81)
Q 3019112919

How many terms of the G.P. 3 , 3/2 , 3/4 , ............. are needed to give the sum (3069)/(512)

Solution:

Let n be the number of terms needed. Given that a = 3, r =1/2 and S_n = (3069)/(512)

Since S_n = (a (1-r^n))/(1-r)

Therefore (3069)/(512) = ( 3 ( 1-1/(2^n)))/(1-1/2) = 6 (1-1/(2^n))

(3069)/(3072) = (1-1/(2^n))

or 1/(2^n) = 1-3069/3072 = 3/3072 = 1/1024

or 2^n = 1024 = 2^n = 2^(10)

which gives n = 10.
Q 3029223111

The sum of first three terms of a G.P. is 13/12 and their product is – 1. Find the common ratio and the terms.

Solution:

Let a/r , a , ar  be the first three terms of the G.P. Then

a/r +ar +a = 13/12 ..............(1)

and (a/r) (a) (ar) = 1 .................(2)

From (2), we get a_3 = – 1, i.e., a = – 1 (considering only real roots)
Substituting a = –1 in (1), we have

 -1/r -1-r = 13/12 or 12r^2+25r+12 = 0

This is a quadratic in r, solving, we get r= -3/4  or -4/3

Thus, the three terms of G.P. are 4/3 , -1 , 3/4 for r = (-3)/4 and r = (-3)/4 and 3/4 , -1 , 4/3 for r = (-4)/3
Q 3069223115

Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.

Solution:

This is not a G.P., however, we can relate it to a G.P. by writing the terms as

S_n = 7 + 77 + 777 + 7777 + ...  to n terms

 = 7/9 [ 9+99+999+9999+........+ to n term ]

 = 7/9 [ (10-1) + (10^2-1) +(10^3-1) +(10^4-1) +......+ n terms]

 = 7/9 [ (10+10^2+10^3 + ............. n terms ) - (1+1+1+......n terms) ]

 = 7/9 [ (10(10^n -1))/(10-1) -n] =7/9 [ (10(10^n-1))/9-n]
Q 3079323216

A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.

Solution:

Here a = 2, r = 2 and n = 10

Using the sum formula S_n = ( a (r^n -1))/(r-1)

we have S_(10) = 2(2^(10)-1) = 2046

Hence, the number of ancestors preceding the person is 2046.

### Geometric Mean (G .M.)

\color{fuchsia} \(★ mathbf\ul"Geometric Mean (G .M.) ")

The geometric mean of two positive numbers a and b is the number color(red)(sqrt(ab )).

Therefore, the geometric mean of 2 and 8 is 4.

We observe that the three numbers 2,4,8 are consecutive terms of a G.P. This leads to a generalisation of the concept of geometric means of two numbers.

Given any two positive numbers a and b, we can insert as many numbers as we like between them to make the resulting sequence in a G.P.

Let color(blue)(G_1, G_2,…, G_n) be n numbers between positive numbers a and b such that color(blue)(a,G_1,G_2,G_3,…,G_n,b) is a G.P.

Thus, b being the (n + 2)th term,we have color(blue)(b = ar^(n + 1)) , or color(blue)(r=(b/a)^(1/(n+1)))

Hence color(blue)(G_1= ar=a(b/a)^(1/(n+1)))

color(blue)(G_2= ar^2=a(b/a)^(2/(n+1)))

color(blue)(G_3= ar^3=a(b/a)^(3/(n+1)))

color(blue)(G_n=ar^n=a(b/a)^(n/(n+1)))
Q 3079423316

Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.

Solution:

Let G_1, G_2,G_3 be three numbers between 1 and 256 such that 1, G_1,G_2,G_3 ,256 is a G.P.

Therefore 256 = r^4 giving r = ± 4 (Taking real roots only)
For r = 4, we have G_1 = ar = 4, G_2 = ar^2 = 16, G_3 = ar^3 = 64
Similarly, for r = – 4, numbers are – 4,16 and – 64.
Hence, we can insert, 4, 16, 64 or – 4, 16, –64, between 1 and 256 so that the resulting
sequences are in G.P.

### Relationship Between A.M. and G.M.

Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.

Then color(blue)(A= (a+b)/2) and color(blue)(G=sqrt(ab))

Thus, we have color(blue)(A – G)=(a+b)/2-sqrt(ab)= (a+b-2sqrt(ab))/2

 \ \ \ \ \ \ \ \ \ \ \ color(blue)(A – G)=(sqrt(a)-sqrt(b))^2/2 >= 0

we obtain the relationship color(red)(A ≥ G.)
Q 3019523419

If A.M. and G.M. of two positive numbers a and b are 10 and 8, respectively, find the numbers.

Solution:

Given that A. M. = (a+b)/2 = 10 ................(1)
and G. M. = sqrt(ab) = 8 ........(2)

From (1) and (2), we get

a + b = 20 ... (3)
ab = 64 ... (4)

Putting the value of a and b from (3), (4) in the identity (a – b)^2 = (a + b)^2 – 4ab, we get

(a-b)^2 = 400-256 = 144

a-b = pm12  .........(5)

Solving (3) and (5), we obtain
a = 4, b = 16 or a = 16, b = 4
Thus, the numbers a and b are 4, 16 or 16, 4 respectively.