Maths help please!!!!!!?

Question

insert 3 numbers between 96 and 3/8 so that the five numbers from a geometric sequence.find the inserted number.

Answer 1

We'll let the geometric sequence be {a(n)} and we'll define a(0) = 96. We know a(4) = 3/8. If r is the constant that we multiply to a term to get the new term, i.e. a(n+1) = r * a(n), then:

a(0) = 96

a(1) = 96r

a(2) = 96r^2

a(3) = 96r^3

a(4) = 96r^4 = 3/8

Now we have an equation to find r:

96r^4 = 3/8

r^4 = 1/256 = 1 / 4^4

r = +1/4 or -1/4

Therefore, there are two sequences of three numbers that will make the five numbers a geometric sequence:

a(0) = 96

a(1) = 24

a(2) = 6

a(3) = 3/2

a(4) = 3/8

or

a(0) = 96

a(1) = -24

a(2) = 6

a(3) = -3/2

a(4) = 3/8

Answer 2

A geometric sequence looks like this:

a, ax, ax^2, ax^3, ax^4, ...

If 96 is a and 3/8 is ax^4, then (ax^4)/a = x^4 = (3/8)/96 = .003906.  Then x = .003906^(1/4) = .25.

So each number is .25 times the prior number.  24 = 96*.25, etc.

24, 6, 1.5