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I don't want the answer to this question, rather just whether it is actually possible. So far I have found m = 10 and r = 2 (or alternatively m = 0 and r = -1/2) which gives the consecutive terms listed in the question values of 12, 24 and 48 (or -8, 4 and -2).

But to work out the summation of the first 10 elements (using the formula (a(r^10-1))/(r-1) ) then surely I need to know a. As the question states that the terms are consecutive but doesn't give information about where in the sequence they appear, is it even possible to work out a and hence the summation? If I understand this correctly then a could be 12, but it could as easily be 6 or 3 or . . . . .

NB I shouldn't assume anything - so I can't assume that 2m-8 is the first term.

Alfonsobonso
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2 Answers2

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HINT: solve the System $$2m+4=(2m-8)q$$ and $$5m-2=(2m-8)q^2$$ solving this System we get $$m=0,q=-\frac{1}{2}$$ or $$m=10,q=2$$

Dr. Sonnhard Graubner
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  • In this case, I know q (or the geometric relation r). M is 10 or 0 and r is 2 or -1/2. But if I don't assume that 2m-8 is the first term of the sequence (I've been told I shouldn't assume that) then how do I work out a? – Alfonsobonso Sep 04 '17 at 14:24
  • is your Problem solved now? – Dr. Sonnhard Graubner Sep 04 '17 at 14:24
  • No, I can work out m and r but, without assuming where in the sequence the consecutive terms appear, is it even possible to find a? – Alfonsobonso Sep 04 '17 at 14:26
  • what is $a$ here? – Dr. Sonnhard Graubner Sep 04 '17 at 14:30
  • It's not given. The only thing given is the text of my initial question above: "Given that 2m-8, 2m+4 and 5m-2 are successive terms of a geometric sequence. Find the value of m and thus the summation of the first 10 elements." – Alfonsobonso Sep 04 '17 at 14:35
  • but i have solved this, what questions anymore? – Dr. Sonnhard Graubner Sep 04 '17 at 14:36
  • You need to know the first term, a, to find the summation. But a isn't given in the question, you're only told that the three terms in the question are consecutive. Is it possible to find the summation without assuming that the first term, a, is 2m-8? – Alfonsobonso Sep 04 '17 at 14:53
  • for example for $$m=10$$ and $$q=2$$ you will get $$a_1=12,a_2=24,a_3=48...$$ and so on – Dr. Sonnhard Graubner Sep 04 '17 at 14:57
  • That's just it - you e assumed that 12 is a1. I've expressly asked my tutor about this and whether to assume that the terms given in the question are the first three of the sequence and he told me "no, if that information was needed it would be in the question". As a result, I don't believe you can solve the problem as, given the limited information, 12 could be a1 but it could also a2 or a3. – Alfonsobonso Sep 04 '17 at 15:11
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Let's set $\begin{cases} a=2m-8\\b=2m+4\\c=5m-2\end{cases}$

If you do not assume any order on the terms, unfortunately you have to solve all $6$ systems. This is not difficult since you'll get a quadratic polynomial to solve each time, but this is for sure very tedious.

Here are the solutions computed with a CAS.

$\begin{array}{cc|cc|c} \# & a_1,a_2,a_3 & equations && solutions \\\hline (1) & a,b,c & b=ra & c=rb & \{r=-\frac 12,\ m=0\}\ \{r=2,\ m=10\}\\ (2) & a,c,b & c=ra & b=rc & \{r=\frac{1\pm i\sqrt 5}{3},\ m=\frac{2\mp 4i\sqrt 5}{7}\}\\ (3) & b,a,c & a=rb & c=ra & \{r=\frac{1\pm \sqrt 7}{2},\ m=-4\mp 2\sqrt 7\}\\ (4) & b,c,a & c=rb & a=rc & \{r=\frac{1\pm i\sqrt 5}{2},\ m=\frac{2\pm 4i\sqrt 5}{7}\}\\ (5) & c,a,b & a=rc & b=ra & \{r=\frac{-1\pm \sqrt 7}{3},\ m=-4\mp 2\sqrt 7\}\\ (6) & c,b,a & b=rc & a=rb & \{r=-2,\ m=0\}\ \{r=\frac 12,\ m=10\}\\ \end{array}$

As you can notice only $(1)$ and $(6)$ are nice (since order is reversed between a,b,c the reason is simply inverted).

Seeing how complicated are the other solutions, I guess your exercise implicitly assumed to sum the first terms only for the $4$ simple cases.

zwim
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