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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.

jiten
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2 Answers2

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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.

Servaes
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  • So, if the question had a different starting point of $4$ and ending point of say $13$, then can get the series with distance $=9$. – jiten Jun 28 '19 at 09:27
  • Yes, that would have been fine. – Servaes Jun 28 '19 at 09:27
  • But, why the distance between the last and first term not taken into account, i.e. $7$. – jiten Jun 28 '19 at 09:31
  • I do not understand your question; the argument shows that a geometric sequence cannot start at $-4$ and then have $3$ as its fifteenth term. What part of that is unclear? – Servaes Jun 28 '19 at 09:33
  • So, the answer by @peterwhy is correct when the first endpoint ($-4$) is not taken as part of the geometric sequence. – jiten Jun 28 '19 at 09:42
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For (ii), it is not that $ar^{14} = 3$, but that

$$a+2a+4a+\cdots + 2^{13}a = 3-(-4)$$

where $a$ is the distance between $-4$ and the next inserted point, and the distance doubles with each point.

There are just $14$ intervals between the $15$ points, so the last inserted point and $3$ has a distance of $2^{13}a$.

Then the first term in the sequence is $-4$. And after that:

  • second term is $-4 + a$,
  • third term is $-4 + a + 2a$,
  • $n$th term is $-4 + \sum_{i=0}^{n-2} 2^ia$
peterwhy
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  • Why is that? The terms of a geometric sequence are $a$, $ar$, $ar^2$, etc. Not the partial sums. And why $r=2$? – Servaes Jun 28 '19 at 09:28
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    @Servaes $a$ is the distance between $-4$ and the next inserted point, and the distance doubles with each point. The last inserted point and $3$ has a distance of $2^{13}a$. – peterwhy Jun 28 '19 at 09:30
  • @Servaes The comment by peterwhy is correct, as actually that was what I meant, but am confused a lot now to find out what works. – jiten Jun 28 '19 at 09:32
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    @jiten But then the roots of the equation together with the $13$ points do not form a geometric sequence; the lengths of the intervals do. – Servaes Jun 28 '19 at 09:35
  • @peterwhy I had interpreted the question differently. My previous comment was directed more at the OP though. At least I understand your answer now :) – Servaes Jun 28 '19 at 09:37
  • Please vet the calculation to get $3$ as last point, as now the sum is $\frac{a(r^{14}-1)}{2-1}=\frac{a(2^{14}-1)}{2-1}$. So, $a(2^{14}-1)=7\implies a = \frac{7}{2^{14}-1}$. This leads to $\frac{\frac{7}{2^{14}-1}(2^{14}-1)}{2-1}=7$. – jiten Jun 28 '19 at 09:51
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    @jiten Did you $-4$? – peterwhy Jun 28 '19 at 09:54