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, G_{1}, G_{2}, ..., G_{n}, b are in G. P.,
then G_{1}, G_{2}, ..., G_{n } are called n G. M.'s between a and b.
The geometric mean of n numbers is defined as the n^{th }root of their product.
Thus if a_{1}, a_{2}, ..., a_{n} are n numbers, then their
G. M. = (a_{1}, a_{2}, ... a_{n})^{1/n}
Let G be the G. M. between a and b, then a, G, b are in G. P.
or, G^{2} =ab
or G = √ab
∴
Given any two positive numbers a and b, any number of geometric means can be inserted a_{1}, a_{2}, a_{3} ..., a_{n }be n geometric means between a and b.
then, a_{1}, a_{2}, a_{3} ..., a_{n },b is a G.P.
Thus, b being the (n + 2)^{th} term, we have
b = ar^{n+1}
or, r^{n+1} = b/a
or,
Hence,
....... .......
...... .......
Further we can show that the product of these n G. M.'s is equal to n^{th} power of the single geometric mean between a and b.
Multiplying a_{1}, a_{2}, ... a_{n}, 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 G_{1}, G_{2}, G_{3}, be the three geometric means between 1 and 256.
Then 1, G_{1}, G_{2}, G_{3}, 256 are in G. P.
If r be the common ratio, then t_{5} = 256
i.e, ar^{4} = 256 = 1
r^{4} = 256
or, r^{2} = 16
or r = ± 4
When r = 4, G_{1} = 1. 4 = 4, G_{2} = 1. (4)^{2} = 16 and G_{3 }= 1. (4)^{3} = 64
When r = – 4, G_{1} = – 4, G_{2} = (1) (–4)^{2} = 16 and G_{3} = (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 = a^{1/2.}b^{1/2}
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