Three consecutive terms of a geometric sequence are $x - 3$, 6 and $x + 2$.

Find the possible values of $x$.

METHOD 1

valid approach     (M1)

eg$\,\,\,\,\,$$r = \frac{6}{{x - 3}},{\text{ }}(x - 3) \times r = 6,{\text{ }}(x - 3){r^2} = x + 2$

correct equation in terms of $x$ only     A1

eg$\,\,\,\,\,$$\frac{6}{{x - 3}} = \frac{{x + 2}}{6},{\text{ }}(x - 3)(x + 2) = {6^2},{\text{ }}36 = {x^2} - x - 6$

correct working     (A1)

eg$\,\,\,\,\,$${x^2} - x - 42,{\text{ }}{x^2} - x = 42$

valid attempt to solve their quadratic equation     (M1)

eg$\,\,\,\,\,$factorizing, formula, completing the square

evidence of correct working     (A1)

eg$\,\,\,\,\,$$(x - 7)(x + 6),{\text{ }}\frac{{1 \pm \sqrt {169} }}{2}$

$x = 7,{\text{ }}x =  - 6$     A1     N4

METHOD 2 (finding r first)

valid approach     (M1)

eg$\,\,\,\,\,$$r = \frac{6}{{x - 3}},{\text{ }}6r = x + 2,{\text{ }}(x - 3){r^2} = x + 2$

correct equation in terms of $r$ only     A1

eg$\,\,\,\,\,$$\frac{6}{r} + 3 = 6r - 2,{\text{ }}6 + 3r = 6{r^2} - 2r,{\text{ }}6{r^2} - 5r - 6 = 0$

evidence of correct working     (A1)

eg$\,\,\,\,\,$$(3r + 2)(2r - 3),{\text{ }}\frac{{5 \pm \sqrt {25 + 144} }}{{12}}$

$r =  - \frac{2}{3},{\text{ }}r = \frac{3}{2}$    A1

substituting their values of $r$ to find $x$     (M1)

eg$\,\,\,\,\,$$(x - 3)\left( {\frac{2}{3}} \right) = 6,{\text{ }}x = 6\left( {\frac{3}{2}} \right) - 2$

$x = 7,{\text{ }}x =  - 6$    A1     N4

[6 marks]

Nearly all candidates attempted to set up an expression, or pair of expressions, for the common ratio of the geometric sequence. When done correctly, these expressions led to a quadratic equation which was solved correctly by many candidates.