Show that the common ratio is \(\frac{1}{2}\) .

Find the value of the eleventh term.

Find the sum of the first eight terms.

Find the number of terms in the sequence for which the sum first exceeds \(2047.968\).

Find the value of the eleventh term.

The sum of the first \(n\) terms of this sequence is \(\frac{n}{2}(3n - 1)\).

(i)     Find the sum of the first 100 terms in this arithmetic sequence.

(ii)    The sum of the first \(n\) terms is \(477\).

     (a)     Show that \(3{n^2} - n - 954 = 0\) .

     (b)     Using your graphic display calculator or otherwise, find the number of terms, \(n\) .

\(1024{r^3} = 128\)     (M1)

\({r^3} = \frac{1}{8}\) or \(r = \sqrt[3]{{\frac{1}{8}}}\)     (M1)

\(r = \frac{1}{2}{\text{ }}(0.5)\)     (AG)

Notes: Award at most (M1)(M0) if last line not seen. Award (M1)(M0) if \(128\) is found by repeated multiplication (division) of \(1024\) by \(0.5\) \((2)\).

[2 marks]

\(1024 \times {0.5^{10}}\)     (M1)

Notes: Award (M1) for correct substitution into correct formula. Accept an equivalent method.

1     (A1)(G2)

[2 marks]

\({S_8} = \frac{{1024\left( {1 - {{\left( {\frac{1}{2}} \right)}^8}} \right)}}{{1 - \frac{1}{2}}}\)     (M1)(A1)

Note: Award (M1) for substitution into the correct formula, (A1) for correct substitution.

OR

(A1) for complete and correct list of eight terms     (A1)

(M1) for their eight terms added     (M1)

\({S_8} = 2040\)     (A1)(G2)

[3 marks]

\(\frac{{1024\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} > 2047.968\)     (M1)(M1)(ft)

Notes: Award (M1) for correct substitution into the correct formula for the sum, (M1) for comparing to \(2047.968\) . Accept equation. Follow through from their expression for the sum used in part (c).

OR

If a list is used: \({S_{15}} = 2047.9375\)     (M1)

\({S_{16}} = 2047.96875\)     (M1)

\(n = 16\)     (A1)(ft)(G2)

Note: Follow through from their expression for the sum used in part (c).

[3 marks]

\({\text{common difference}} = 3\) (may be implied)     (A1)

\({u_{11}} = 31\)     (A1)(G2)

[2 marks]

(i)     \(\frac{{100}}{2}(3 \times 100 - 1)\)     OR     \(\frac{{100(2 + 99 \times 3)}}{2}\)     (M1)

         \(14 950\)     (A1)(G2)

(ii)     (a)     \(\frac{n}{2}(3n - 1) = 477\)     OR     \(\frac{n}{2}(2 + 3(n - 1)) = 477\)     (M1)

                   \(3{n^2} - n = 954\)     (M1)

                   \(3{n^2} - n - 954 = 0\)     (AG)

Notes: Award second (M1) for correct removal of denominator or brackets and no further incorrect working seen. Award at most

(M1)(M0) if last line not seen.

        (b)     \(18\)     (G2)

Note: If both solutions to the quadratic equation are seen and the correct value is not identified as the required answer, award (G1)(G0).

[6 marks]

Part A: Geometric sequences/series

The majority of the candidates were not able to offer a satisfactory justification in a) and only scored 1 mark.

Part A: Geometric sequences/series

Parts b) and c) were mostly well answered.

Part A: Geometric sequences/series

Parts b) and c) were mostly well answered.

Part A: Geometric sequences/series

The responses to part d) were often weak. Those candidates who set up the equation scored two marks but very few of them were able to reach the correct final answer.

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly.

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly. Parts b)(ii)a) and b)(ii)b) were poorly answered. Many candidates did not know how to approach the “show that” question. A few were able to solve the quadratic equation using the GDC. Those who attempted to solve it without the GDC generally failed to find the correct answer.