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

Jul 26, 2016

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

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

$m = \frac{7}{8} \pm \frac{\sqrt{15}}{8} i$

#### Explanation:

If $a , b , c$ is a geometric sequence, then $\frac{b}{a} = \frac{c}{b}$ and hence ${b}^{2} = a c$.

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

${\left(3 m - 2\right)}^{2} = \left(m - 1\right) \left(5 m\right)$

which expands to:

$9 {m}^{2} - 12 m + 4 = 5 {m}^{2} - 5 m$

Subtract $5 {m}^{2} - 5 m$ from both sides to get:

$4 {m}^{2} - 7 m + 4 = 0$

The discriminant $\Delta$ of a quadratic $a {x}^{2} + b x + c$ is given by the formula:

$\Delta = {b}^{2} - 4 a c$

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

$\Delta = {\left(- 7\right)}^{2} - 4 \left(4\right) \left(4\right) = 49 - 64 = - 15$

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

$m = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{7 \pm \sqrt{15} i}{8}$

$= \frac{7}{8} \pm \frac{\sqrt{15}}{8} i$

These values lead to geometric sequences of Complex numbers.