Insert $13$ real numbers between the roots of the equation: $x^2 +x−12 = 0$ in a few ways that these $13$ numbers together with the roots of the equation will form the first $15$ elements of a sequence. Write down in an explicit form the general (nth) element of the formed sequence.
Both roots of $x^2 +x−12 = 0$ are in reals as $D= 49$, these are: $x = \frac{-1 \pm 7}{2}= 3, -4$.
i) Form an arithmetic sequence, i.e. the distance between the terms is the same.
Insert $13$ reals between these two in equidistant manner.
As the distance is $7$, so, need equal intra-distance $=\frac {7}{14}$.
So, the first term is at $-4$, next at $-4+\frac {7}{14}=\frac {-45}{14}$, & so on.
ii) Make the distance double with each next point, i.e. there is a g.p. of the minimum distance.
Let the first term be $a$, common ratio term be $r=2$, & $\,2^{14}r\,$ is the maximum gap between the consecutive terms.
The sum of the geometric series is given by:
$a+ar+ar^2+\cdots+ar^{14}$, or
$a+2a+4a+8a+16a+\cdots+2^{14}a = a\frac{2^{15}-1}{2-1}=a(2^{15}-1)$
The last term $\,ar^{14}=3\implies a= \frac{3}{r^{14}} = \frac{3}{2^{14}}.$
So, the series starts at the second point (i.e., the one after $-4$).
This second (starting) point is at : $-4+\frac{3}{2^{14}}$, third point at : $-4+3\frac{3}{2^{14}}$, fourth point at : $-4+7\frac{3}{2^{14}}$,
The last point should act as a check, as its value is $\,3\,$ giving us $-4+\frac{3}{2^{14}}(2^{15}-1)$, which should equal $3$, but is not leading to that.
There is no real geometric series with first term $-4$ and fifteenth term $3$; if the ratio between the terms is $r$ then $a=-4$ and $ar^{14}=3$ and so $r^{14}=\tfrac3a=-\tfrac34$. But $r^{14}\geq0$ because $r$ is a real number, a contradiction.