Question 529087
<pre>
The other tutor's solution is incorrect.

x-3, x+1, 2x+8

Let r = the common ratio 

Then we have the system of two equations in two unknowns:

r(x-3) = x+1
r(x+1) = 2x+8

Solving each for r:

r = {{{(x+1)/(x-3)}}}
r = {{{(2x+8)/(x+1)}}}

Setting the right sides equal to each other, since both
equal to r:

{{{(x+1)/(x-3)}}} = {{{(2x+8)/(x+1)}}}

Cross-multiplying:

(x+1)(x+1) = (x-3)(2x+8)

x² + 2x + 1 = 2x² + 2x - 24

0 = x² - 25

0 = (x - 5)(x + 5)

 x - 5 = 0   x + 5 = 0
     x = 5       x = -5

As they told us, x = 5 is one of the values and it
makes the sequence

x-3, x+1, 2x+8 become

5-3, 5+1, 2(5)+8

2, 6, 18

and the common ratio is {{{6/2}}} = {{{18/6}}} = 3.

The other value of x is -5.  It makes the sequence

x-3, x+1, 2x+8 become:

-5-3, -5+1, 2(-5)+8

-8, -6, -2

and the ratio is {{{(-6)/(-8)}}} = {{{(-2)/(-6)}}} = {{{1/3}}}  

Since the common ratio is less than 1, we can sum the series
to infinity with the equation:

{{{S[infinity]}}} = {{{a[1]/(1-r)}}}

where {{{a[1]}}} is the first term -8, and r = {{{1/3}}}

Substituting:

{{{S[infinity]}}} = {{{(-8)/(1-(1/3))}}} = {{{(-8)/(2/3)}}} = {{{-8*expr(3/2)}}} = -12.

Edwin</pre>