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)