The sum of the first three terms of a geometric sequence is \(62.755\), and the sum of the infinite sequence is \(440\). Find the common ratio.

correct substitution into sum of a geometric sequence     A1

eg   \(62.755 = {u_1}\left( {\frac{{1 - {r^3}}}{{1 - r}}} \right)\) , \({u_1} + {u_1}r + {u_1}{r^2} = 62.755\)

correct substitution into sum to infinity     A1

eg   \(\frac{{{u_1}}}{{1 - r}} = 440\)

attempt to eliminate one variable     (M1)

eg substituting \({u_1} = 440(1 - r)\)

correct equation in one variable     (A1)

eg   \(62.755 = 440(1 - r)\left( {\frac{{1 - {r^3}}}{{1 - r}}} \right)\) , \(440(1 - r)(1 + r + {r^2}) = 62.755\)

evidence of attempting to solve the equation in a single variable     (M1)

eg sketch, setting equation equal to zero, \(62.755 = 440(1 - {r^3})\)

\(r =0.95 = \frac{{19}}{{20}}\)     A1     N4

[6 marks]

Many candidates were able to successfully obtain two equations in two variables, but far fewer were able to correctly solve for the value of \(r\). Some candidates had misread errors for either \(440\) or \(62.755\), with some candidates taking the French and Spanish exams mistaking the decimal comma for a thousands comma. Many candidates who attempted to solve algebraically did not cancel the \(1 - r\) from both sides and ended up with a 4^th degree equation that they could not solve. Some of these candidates obtained the extraneous answer of \(r - 1\) as well. Some candidates used a minimum of algebra to eliminate the first term and then quickly solved the resulting equation on their GDC.