The sum of the first two terms of a geometric series is 10 and the sum of the first four terms is 30.

(a)     Show that the common ratio \(r\) satisfies \({r^2} = 2\).

(b)     Given \(r = \sqrt 2 \)

          (i)     find the first term;

          (ii)     find the sum of the first ten terms.

(a)     METHOD 1

\(a + ar = 10\)     A1

\(a + ar + a{r^2} + a{r^3} = 30\)     A1

\(a + ar = 10 \Rightarrow a{r^2} + a{r^3} = 10{r^2}\)     or     \(a{r^2} + a{r^3} = 20\)     M1

\(10 + 10{r^2} = 30\)     or     \({r^2}(a + ar) = 20\)     A1

\( \Rightarrow {r^2} = 2\)     AG

METHOD 2

\(\frac{{a(1 - {r^2})}}{{1 - r}} = 10\) and \(\frac{{a(1 - {r^4})}}{{1 - r}} = 30\)     M1A1

\( \Rightarrow \frac{{1 - {r^4}}}{{1 - {r^2}}} = 3\)     M1

leading to either \(1 + {r^2} = 3{\text{   (or }}{r^4} - 3{r^2} + 2 = 0)\)   A1

\( \Rightarrow {r^2} = 2\)     AG

[4 marks]

(b)     (i)     \(a + a\sqrt 2  = 10\)

\( \Rightarrow a = \frac{{10}}{{1 + \sqrt 2 }}\)   or   \(a = 10\left( {\sqrt 2  - 1} \right)\)     A1

(ii)     \({S_{10}} = \frac{{10}}{{1 + \sqrt 2 }}\left( {\frac{{{{\sqrt 2 }^{10}} - 1}}{{\sqrt 2  - 1}}} \right){\text{ }}( = 10 \times 31)\)     M1

\( = 310\)     A1

[3 marks]

Total [7 marks]

This question was invariably answered very well. Candidates showed some skill in algebraic manipulation to derive the given answer in part a). Poor attempts at part b) were a rarity, though the final mark was sometimes lost after a correctly substituted equation was seen but with little follow-up work.