Geometric mean the special type of average.
How to find geometric mean of two numbers ?
Let us consider a and b.
To find geometric mean of a and b, use the formula
= √(a ⋅ b)
How to find geometric mean of three numbers ?
Let us consider a, b and c.
To find geometric mean of a, b and c, use the formula
∛(a ⋅ b ⋅ c)
Note :
we multiply the numbers altogether and take the n^{th} root of the multiplied numbers, where n is the total number of data values.
Problem 1 :
Find the geometric mean of the following numbers
a) 5 and 25
b) 7 and 63
c) -2 and -8
Solution :
a) 5 and 25
Geometric mean = √(x_{1} ⋅ x_{2})
= √(5 ⋅ 25)
= √125
= √(25 × 5)
= 5√5
So, geometric mean is 5√5.
b) 7 and 63
Geometric mean = √(x_{1} ⋅ x_{2})
= √(7 ⋅ 63)
= √441
= 21
So, geometric mean is 21.
c) -2 and -8
Geometric mean = √(x_{1} ⋅ x_{2})
= √((-2) ⋅ (-8))
= √16
= 4
So, geometric mean is 4.
Problem 2 :
Insert 3 numbers between 4 and 324 such that the resulting sequence in a G. P.
Solution :
Let the numbers be G_{1}, G_{2}, and G_{3}.
4, G_{1}, G_{2}, G_{3}, 324 are in G.P.
T_{5} = 324
t_{n} = ar^{(n - 1)}
ar^{4} = 324
4 × r^{4} = 324
Dividing 4 on both sides.
(4 × r^{4})/4 = 324/4
r^{4} = 81
r^{4} = 3^{4}
r = 3
G_{1} = 4 × 3 = 12
G_{2} = 4 × 3^{2} = 36
G_{3} = 4 × 3^{3} = 108
So, the numbers be 12, 36, 108.
Problem 3 :
A G.P is given 1/2, 1, 2, 4, … If a constant k is multiplied in each term of the G.P., then find the 10^{th} term of the G.P.
Solution :
Accordingly the properties of geometric progression, if every term is multiplied by some non zero constant, then the new progression created is also a geometric progression with same common ratio.
To find the10^{th} term of the G.P :
.a_{n }= a_{1}r^{n - 1}
a_{10 }= 1/2 × (2)^{10 - 1}
= 1/2 × 2^{9}
= 1/2 × 256
a_{10 }= 128
So, 10th term of the G.P is a_{10 }= 128.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM