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.

Question

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.

Answer 1

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 $$m=10,q=2$$

Answer 2

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.