Write down an expression for the common ratio, $r$.

Hence, show that $m$ satisfies the equation ${m^2} + 3m - 40 = 0$.

Find the two possible values of $m$.

Find the possible values of $r$.

The sequence has a finite sum.

State which value of $r$ leads to this sum and justify your answer.

The sequence has a finite sum.

Calculate the sum of the sequence.

correct expression for $r$     A1     N1

eg   $r = \frac{6}{{m - 1}},{\text{ }}\frac{{m + 4}}{6}$

[2 marks]

correct equation     A1

eg     $\frac{6}{{m - 1}} = \frac{{m + 4}}{6},{\text{ }}\frac{6}{{m + 4}} = \frac{{m - 1}}{6}$

correct working     (A1)

eg     $(m + 4)(m - 1) = 36$

correct working     A1

eg     ${m^2} - m + 4m - 4 = 36,{\text{ }}{m^2} + 3m - 4 = 36$

${m^2} + 3m - 40 = 0$     AG     N0

[2 marks] 

valid attempt to solve     (M1)

eg     $(m + 8)(m - 5) = 0,{\text{ }}m = \frac{{ - 3 \pm \sqrt {9 + 4 \times 40} }}{2}$

$m =  - 8,{\text{ }}m = 5$     A1A1     N3

[3 marks]

attempt to substitute any value of $m$ to find $r$     (M1)

eg     $\frac{6}{{ - 8 - 1}},{\text{ }}\frac{{5 + 4}}{6}$

$r = \frac{3}{2},{\text{ }}r =  - \frac{2}{3}$     A1A1     N3

[3 marks]

$r =  - \frac{2}{3}$   (may be seen in justification)     A1

valid reason     R1     N0

eg     $\left| r \right| < 1,{\text{ }} - 1 < \frac{{ - 2}}{3} < 1$

Notes: Award R1 for $\left| r \right| < 1$ only if A1 awarded.

[2 marks]