Question 1057668
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The sum of the second and third terms of a geometric progression is six times the fourth term. 
Find the two possible values of the common ratio.
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The condition says

{{{ar + ar^2}}} = {{{6ar^3}}}

where "a" is the first term and "r" is the common ratio.
It leads to the equation

{{{6r^2 - r - 1}}} = 0,  or, in the factorized form

(2r-1)*(3r+1) = 0.

The roots are {{{1/2}}} and {{{-1/3}}}.

It is your answer.
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