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Mathematics · Ch 8 — Sequences and Series

Geometric Mean (G.M.)

8.4.3

Geometric Mean (G.M.)

The Geometric Mean of Two Numbers

For any two positive numbers aa and bb, their geometric mean (G.M.) is defined as ab\sqrt{ab}.

For example, the geometric mean of 22 and 88 is 2×8=16=4\sqrt{2 \times 8} = \sqrt{16} = 4.

What makes this number special? Notice that 2,4,82, 4, 8 are three consecutive terms of a geometric progression (G.P.) with common ratio 22. This is not a coincidence — it points to a deeper idea: the geometric mean of two numbers is the number that sits exactly between them in a G.P.

Important

If aa and bb are positive, then aa, ab\sqrt{ab}, bb are in G.P.

Check: aba=bab=ba\frac{\sqrt{ab}}{a} = \frac{b}{\sqrt{ab}} = \sqrt{\frac{b}{a}}, so the common ratio is b/a\sqrt{b/a}.


Inserting Multiple Geometric Means Between Two Numbers

The idea extends naturally. Given any two positive numbers aa and bb, we can insert as many numbers as we like between them so that the entire sequence — including aa and bb — forms a G.P.

Let G1,G2,G3,…,GnG_1, G_2, G_3, \dots, G_n be nn numbers inserted between aa and bb such that

a,G1,G2,G3,…,Gn,ba, G_1, G_2, G_3, \dots, G_n, b

is a G.P.

Here aa is the first term, and bb is the (n+2)(n+2)-th term. If the common ratio is rr, then

b=ar(n+2)−1=arn+1.b = a r^{(n+2)-1} = a r^{n+1}.

From this we get

rn+1=ba⇒r=(ba)1n+1.r^{n+1} = \frac{b}{a} \quad \Rightarrow \quad r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}}.

Now each inserted geometric mean can be written explicitly:

G1=ar=a(ba)1n+1,G2=ar2=a(ba)2n+1,G3=ar3=a(ba)3n+1,  ⋮Gn=arn=a(ba)nn+1.\begin{aligned} G_1 &= a r = a \left(\frac{b}{a}\right)^{\frac{1}{n+1}}, \\[4pt] G_2 &= a r^2 = a \left(\frac{b}{a}\right)^{\frac{2}{n+1}}, \\[4pt] G_3 &= a r^3 = a \left(\frac{b}{a}\right)^{\frac{3}{n+1}}, \\[4pt] &\ \ \vdots \\[4pt] G_n &= a r^n = a \left(\frac{b}{a}\right)^{\frac{n}{n+1}}. \end{aligned}

Note

The kk-th geometric mean (counting from aa) is Gk=a(ba)kn+1G_k = a \left(\frac{b}{a}\right)^{\frac{k}{n+1}}, for k=1,2,…,nk = 1, 2, \dots, n.


Worked Example: Inserting Three Numbers

Example 12. Insert three numbers between 11 and 256256 so that the resulting sequence is a G.P.

Solution. Let the three numbers be G1,G2,G3G_1, G_2, G_3. We want

1,G1,G2,G3,2561, G_1, G_2, G_3, 256

to be a G.P. Here a=1a = 1, b=256b = 256, and n=3n = 3 (three inserted terms). The total number of terms is 55, so bb is the 55-th term:

256=1⋅r4⇒r4=256.256 = 1 \cdot r^{4} \quad \Rightarrow \quad r^4 = 256.

Taking real roots only, r=±4r = \pm 4.

Case 1: r=4r = 4

G1=1⋅4=4,G2=1⋅42=16,G3=1⋅43=64.G_1 = 1 \cdot 4 = 4, \quad G_2 = 1 \cdot 4^2 = 16, \quad G_3 = 1 \cdot 4^3 = 64.

So the sequence is 1,4,16,64,2561, 4, 16, 64, 256.

Case 2: r=−4r = -4

G1=1⋅(−4)=−4,G2=1⋅(−4)2=16,G3=1⋅(−4)3=−64.G_1 = 1 \cdot (-4) = -4, \quad G_2 = 1 \cdot (-4)^2 = 16, \quad G_3 = 1 \cdot (-4)^3 = -64. …