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

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Answer 1 · 2016-07-26T08:39:31

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

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

$m = \frac{7}{8} \pm \frac{\sqrt{15}}{8} i$ $m = 7/8+-sqrt(15)/8i$

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

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

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

which expands to:

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

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

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

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.

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