# How do you find k such that k+1, 4k, 3k+5 is a geometric sequence?

Dec 5, 2015

$k = 1$ or $k = - \frac{5}{13}$

#### Explanation:

If $k + 1 , 4 k , 3 k + 5$ is a geometric sequence
then the ratio between successive terms is equal.

$\frac{k + 1}{4 k} = \frac{4 k}{3 k + 5}$

$\Rightarrow \left(k + 1\right) \left(3 k + 5\right) = {\left(4 k\right)}^{2}$

$\Rightarrow 3 {k}^{2} + 8 k + 5 = 16 {k}^{2}$

$\Rightarrow 13 {k}^{2} - 8 k - 5 = 0$

We might be able to factor this directly or we could use the quadratic formula to determine the roots:
color(white)("XXX")k= (8+-sqrt((-8)^2-4(13)(-5)))/(2(13)

$\textcolor{w h i t e}{\text{XXXX}} = \frac{8 \pm \sqrt{324}}{2 \left(13\right)}$

color(white)("XXXX")= (8+-sqrt(324))/(2(13)

$\textcolor{w h i t e}{\text{XXX}} = \frac{8 \pm 18}{2 \left(13\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{4 \pm 9}{13}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{13}{13} = 1$ or $= - \frac{5}{13}$

We could (and probably should) verify these results by checking that for each of these values of $k$ the given sequence is geometric.

If $k = 1$
then $k + 1 , 4 k , 3 k + 5$
becomes $2 , 4 , 8$ with an obvious common ratio of $2$

If $k = - \frac{5}{13}$
then $k + 1 , 4 k , 3 k + 5$
becomes (with a little more effort) $\frac{8}{13} , - \frac{20}{13} , \frac{50}{13}$
with a common ratio of $\left(- \frac{5}{2}\right)$