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Lesson Posted on 06/09/2018 Learn Progressions
Raj Kumar
I am Six Sigma Black belt trained from American Society of Quality I am 2011 pass out in B.tech from...
If a is the first term and r is the common ratio of the geometric progression,
then its nth term is given by an = arn-1
The sum Sn of the first n terms of the G.P. is given by
Sn = a (rn – 1)/ (r-1), when r ≠1; = na if r =1
If -1 < x < 1, then limxn = 0, as n →∞. Hence, the sum of an infinite G.P. is 1+x+x2+ ….. = 1/(1-x)
If -1 < r< 1, then the sum of the infinite G.P. is a +ar+ ar2+ ….. = a/(1-r)
If each term of the G.P is multiplied or divided by a non-zero fixed constant, the resulting sequence is again a G.P.
If a1, a2, a3, …. andb1, b2, b3, … are two geometric progressions, then a1b1, a2b2, a3b3, …… is also a geometric progression and a1/b1, a2/b2, ... ... ..., an/bn will also be in G.P.
Suppose a1, a2, a3, ……,an are in G.P. then an, an–1, an–2, ……, a3, a2, a1 will also be in G.P.
Taking the inverse of a G.P. also results a G.P. Suppose a1, a2, a3, ……,an are in G.P then 1/a1, 1/a2, 1/a3 ……, 1/an will also be in G.P
If we need to assume three numbers in G.P. then they should be assumed as a/b, a, ab (here common ratio is b)
Four numbers in G.P. should be assumed as a/b3, a/b, ab, ab3 (here common ratio is b2)
Five numbers in G.P. a/b2, a/b, a, ab, ab2 (here common ratio is b)
If a1, a2, a3,… ,an is a G.P (ai> 0 ∀i), then log a1, log a2, log a3, ……, log an is an A.P. In this case, the converse of the statement also holds good.
If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b
= √ac is the geometric mean of a and c.
read lessLesson Posted on 23/08/2018 Learn Progressions
Raj Kumar
I am Six Sigma Black belt trained from American Society of Quality I am 2011 pass out in B.tech from...
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Lesson Posted on 30/05/2018 Learn Progressions
Geometric Progression : Basics
Hemant P.
1. I have taught at various prestigious institutes of repute in India and abroad. 2. Worked at (i)...
Geometric Mean (GM)
The geometric mean is similar to the arithmetic mean, but it uses the property of multiplication, to find the mean between any two numbers.
GM of two numbers a and b may be defined as G = √(ab)
For example, GM of 2 and 3 is √6.
We can find a general formula for inserting a GM between any two arbitrary numbers a and b.
If G is the geometric mean between a and b, then by definition of GP a, G, b are in GP.
Or G2=a.b
We can also find a formula for inserting n GM in between two numbers a and b.
Let G1, G2, G3 …. Gn be then GMs between a and b.
Now a, G1, G2, G3…. Gn, b are in GP.
By definition of the last term of a GP b = (n+2)th term of GP.
If R=common ratio, then by definition of GP, b = a.Rn+1
Now G1=a?R
G2=a?R2
G3=a?R3
and Gn=a?Rn
We will take a few examples of GM too.
Find the GM of the following and insert n geometric means between them.
e and 1
7 and tan x
log x and sec-1x
Solution:
(i) The GM between e and 1 is e. To insert n GMs, we use the definition of geometric progressions.
Let G1, G2, G3, … Gn, be n geometric means inserted between e and 1.
Also, let common ratio be R.
By the definition of GP = (n+2)th term of the GP.
1 =eRn+1
Or R= (1/e)1/(n+1)
Hence G1 = e⋅R = e⋅(1/e)1/(n+1)
G2=e⋅R2=e⋅(1/e)2/(n+1)
G3=e⋅R=e⋅(1/e)3/(n+1)
and Gn=e⋅R=e⋅(1/e)n/(n+1)
(ii) Proceeding as before the GM of 7 and tanx is √(7tanx).
Let G1, G2, G3,… Gn, be n geometric means inserted between 7 and tanx.Also, let common ratio be R.
To get the n geometric means, we proceed as before.
Here tanx=(n+2)th term.
tanx=7⋅Rn+1
Or Rn+1= tanx/7
R = (tanx/7)1/n+1
G1 = 7R = 7⋅(tanx/7)1/n+1
G2 = 7R2 = 7⋅(tanx/7)2/n+1
G3 = 7R3 = 7⋅(tanx/7)3/n+1
and Gn=7Rn=7(tanx/7)n/n+1
(iii) Proceeding exactly as before, the GM of logx and sec-1x will be √(logx⋅sec-1x)
Let G1, G2, G3, … Gn, be n geometric means inserted between logx and sec-1x.
Also, let common ratio be R.
To insert n GMs between logx and sec-1x, we take the sequence as
logx, G1, G2, G3,… Gn, sec-1x.
Gn = (n+2)th term = sec-1x = logx.Rn+1
R= (sec-1x/logx)1/(n+1)
Now as before G1=logx⋅R=logx⋅(sec-1x/logx)1/(n+1)
G2=logx⋅R2=logx⋅(sec-1x/logx)2/(n+1)
G3=logx⋅R3=logx⋅(sec-1x/logx)3/(n+1)
and Gn=logx⋅Rn=logx⋅(sec-1x/logx)n/(n+1)
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