If a, G, b are in G. P., then G is called the geometric mean between a and b.
If three numbers are in G. P., the middle one is called the geometric mean between the other two.
If a, G1, G2, ..., Gn, b are in G. P.,
then G1, G2, ..., Gn are called n G. M.'s between a and b.
The geometric mean of n numbers is defined as the nth root of their product.
Thus if a1, a2, ..., an are n numbers, then their
G. M. = (a1, a2, ... an)1/n
Let G be the G. M. between a and b, then a, G, b are in G. P.
or, G2 =ab
or G = √ab
Given any two positive numbers a and b, any number of geometric means can be inserted a1, a2, a3 ..., an be n geometric means between a and b.
then, a1, a2, a3 ..., an ,b is a G.P.
Thus, b being the (n + 2)th term, we have
b = arn+1
or, rn+1 = b/a
Further we can show that the product of these n G. M.'s is equal to nth power of the single geometric mean between a and b.
Multiplying a1, a2, ... an, we have
= (single G. M. between a and b)n
Example 1. Find the G. M. between 3/2 and 27/2.
Solution: We know that if a is the G. M. between a and b, then
G = √ab
Example 2. Insert three geometric means between 1 and 256.
Solution: Let G1, G2, G3, be the three geometric means between 1 and 256.
Then 1, G1, G2, G3, 256 are in G. P.
If r be the common ratio, then t5 = 256
i.e, ar4 = 256 = 1
r4 = 256
or, r2 = 16
or r = ± 4
When r = 4, G1 = 1. 4 = 4, G2 = 1. (4)2 = 16 and G3 = 1. (4)3 = 64
When r = – 4, G1 = – 4, G2 = (1) (–4)2 = 16 and G3 = (1) (–4)3 = –64
∴ G.M. between 1 and 256 are 4, 16, 64, or, – 4, 16, –64.
Example 3. If 4, 36, 324 are in G. P. insert two more numbers in this progression so thatit again forms a G. P.
Solution: G. M. between 4 and
G. M. between 36 and
If we introduce 12 between 4 and 36 and 108 betwen 36 and 324, the numbers 4, 12, 36, 108, 324 form a G. P.
∴ The two new numbers inserted are 12 and 108.
Example 4. Find the value of n such that may be the geometric mean between a and b.
Solution: If x be G. M. between a and b, then
x = a1/2.b1/2