If the terms a, b, and c are in geometric progression (GP) then, the middle term (b) is called the geometric mean (GM) of the other two terms (a and c). Therefore, the Geometric Mean between two terms a and c is given by:
GM (b) =ac [Where, a and c > 0]
The Geometric Mean of as sequence a1,a2,a3,a4,a5,..... up to n terms can be defined as:
If ‘n’ number of Geometric Means G1,G2,G3,G4,...,Gn are inserted between two numbers a and b such that the resulting sequence a, G1, G2, G3, G4, . . . , Gn, b forms the geometric progression, then,
The first term = a and the (n + 2)th term = b
Therefore, G1 =a(ab)n+11
Similarly, G2 =a(ab)n+12
Therefore, Gn =a(ab)n+1n
Also, the product of n geometric means between a and b is equal to the nth power of the single geometric mean between a and b.
i.e. ∑r=1nGr=Gn [where G = single geometric mean between a and b]
Relationship between Arithmetic Mean and Geometric Mean:
1: Arithmetic Mean ≥ Geometric Mean
AM = A = 2a+b. . . . . . . (1)
And, GM = G = ab. . . . . . . . . . (2)
Now Equation (1) – Equation (2) we get,
A – G = 2a+b−ab
i.e. A – G ≥ 0
This implies that A ≥ G.
2: The quadratic equation whose roots are a and b can be written as x2–2Ax+G2=0, [Where A and G are arithmetic and geometric mean between a and b respectively]
If a and b are the roots of the quadratic equation then, x2–(a+b)x+ab=0
Since, a + b = 2A and ab = G2
Therefore, the above equation becomesx2–2Ax+G2=0
3: If A and G are arithmetic and geometric mean between two numbers a and b respectively then
Therefore, by using quadratic formula we get
(a, b) = 22A±4A2−4G2
Or, (a, b) = A±A2−G2
Geometric Mean IIT JEE Problems
Example 1: Insert three geometric means between 2 and 162.
Given a = 2, b = 162, and n = 3
Hence, the 3 geometric means between 2 and 162 are 6, 18, and 54 respectively.
Example 2: Find the geometric Mean of 4 and 36.
Given a = 4 and b = 36
Therefore, the Geometric Mean = ab=4×36=12
Example 3: The arithmetic mean of two positive numbers a and b is 1 more than its Geometric Mean. Find the value of a and b if their difference is 8.
According to the given condition
a – b = 8
Or, (a−b)(a+b)=8. . . . . . (1)
And, AM – GM = 1 (given)
Or, a−b=±2. . . . . . . . (2)
Now on substituting the values of equation (2) in equation (1) we get
a+b=±4. . . . . . . . . . (3)
Now on squaring both LHS and RHS of equation (2) and (3) we get
a+b−2ab=4. . . . . . . . . . . (4)
And, a+b+2ab=16. . . . . . . . . . (5)
On Solving equation (4) and equation (5), we get
ab = 9
so b = 9/a……(6)
And a + b = 10 . . . . . (7)
On substituting the values of b in equation (7), we get
a + 9/a = 10
Now, using quadratic formula
= 9 or 1
Hence, when a = 9, b = 1 and when a = 1, b = 9
Therefore, the required numbers are 9 and 1.
Example 4: If the Arithmetic and Geometric mean of two positive real numbers a and b are 13 and 12, respectively. Find the value of a and b.
Given: A = 13 and G = 12
Note: If A and G are arithmetic and geometric mean between two numbers a and b respectively then:
Therefore, a = 18, and b = 8
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