If #m-1, 3m-2, 5m# is a geometric sequence, then what is the value of #m# ?

1 Answer
Jul 26, 2016

Answer:

There is no Real value of #m# resulting in a geometric sequence.

It is possible to get a geometric sequence of Complex numbers with:

#m = 7/8+-sqrt(15)/8i#

Explanation:

If #a, b, c# is a geometric sequence, then #b/a = c/b# and hence #b^2 = ac#.

So in order for #m-1, 3m-2, 5m# to be a geometric sequence, we must have:

#(3m-2)^2 = (m-1)(5m)#

which expands to:

#9m^2-12m+4 = 5m^2-5m#

Subtract #5m^2-5m# from both sides to get:

#4m^2-7m+4 = 0#

The discriminant #Delta# of a quadratic #ax^2+bx+c# is given by the formula:

#Delta = b^2-4ac#

So in the case of this quadratic in #m# (which has #a=4#, #b=-7#, #c=4#), we find:

#Delta = (-7)^2-4(4)(4) = 49-64 = -15#

Since #Delta < 0# there are no Real zeros. We can find Complex zeros using the quadratic formula:

#m = (-b+-sqrt(b^2-4ac))/(2a)#

#= (-b+-sqrt(Delta))/(2a)#

#= (7+-sqrt(15)i)/8#

#= 7/8+-sqrt(15)/8i#

These values lead to geometric sequences of Complex numbers.