**The Name ’Geometric’**

If *a*, *b*, and *c* are any consecutive terms of a geometric sequence then* b/a=c/b*.

∴*b*^{2}=*ac* and so *b=±√ac* where Mac is the geometric mean of *a* and *c*.

**Geometric Mean**

When three quantities are in G.P., the middle one is called the Geometric Mean (G.M.) between the other two. Thus G will be the G.M. between *a* and *b* if *a*, *G*, *b* are in G.P.

To Find G.M between *a* and *b*:

Let, *G* be the G.M. between *a* and *b*

Then *a*, *G*, *b* are in G.P. ∴

Hence the G.M. between two quantities is equal to the square root of their product.

Example 1. Find the G.M. between 8 and 72.

Solution:

Answer:

*G*=±24

If we consider three consecutive terms in a geometric sequence {*x*, *y*, and *z*} then

where

*r*is the common factor.

Thus, the middle term, *y*, called the geometric mean, can be calculated in terms of the outer two terms, *x* and *z*.

For a geometric sequence {…,

x,y,z, …}

y^{2}=xz

If we need to find three unknown consecutive terms in geometric sequence, we let the terms be:

Ex2. If 5, *x*, 45 are the first three terms of a geometric sequence, determine the value of *x*.

Ex3. Find *k* given that 4, *k*, and *k*^{2}-1 are consecutive terms of a geometric sequence.

Since the first two terms are 4 and *k*, the common ratio must be ¼*k*, the second term divided by the first one. That’s how you find a common ratio. This indicates that the ratio of the third term to the second must share that value. Not only does this give an equation to solve, it’s a proportion! Those are almost fun.

Yes, there’s no reason that

*k*can’t be negative. Feel free to check it if you’d like by writing down the terms. I just thought it through, and I’m convinced. And I don’t want to type any more fractions or radicals for this problem.

Let’s read post •Arithmetic Mean ≥ Geometric Mean ≥ Harmonic Mean (AM ≥ GM ≥ HM)👈.