Date | November 2011 | Marks available | 7 | Reference code | 11N.2.hl.TZ0.12 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 12 | Adapted from | N/A |
Question
In an arithmetic sequence the first term is 8 and the common difference is \(\frac{1}{4}\). If the sum of the first 2n terms is equal to the sum of the next n terms, find n.
If \({a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }} \ldots \) are terms of a geometric sequence with common ratio \(r \ne 1\), show that \({({a_1} - {a_2})^2} + {({a_2} - {a_3})^2} + {({a_3} - {a_4})^2} + \ldots + {({a_n} - {a_{n + 1}})^2} = \frac{{a_1^2(1 - r)(1 - {r^{2n}})}}{{1 + r}}\).
Markscheme
\({S_{2n}} = \frac{{2n}}{2}\left( {2(8) + (2n - 1)\frac{1}{4}} \right)\) (M1)
\( = n\left( {16 + \frac{{2n - 1}}{4}} \right)\) A1
\({S_{3n}} = \frac{{3n}}{2}\left( {2 \times 8 + (3n - 1)\frac{1}{4}} \right)\) (M1)
\( = \frac{{3n}}{2}\left( {16 + \frac{{3n - 1}}{4}} \right)\) A1
\({S_{2n}} = {S_{3n}} - {S_{2n}} \Rightarrow 2{S_{2n}} = {S_{3n}}\) M1
solve \(2{S_{2n}} = {S_{3n}}\)
\( \Rightarrow 2n\left( {16 + \frac{{2n - 1}}{4}} \right) = \frac{{3n}}{2}\left( {16 + \frac{{3n - 1}}{4}} \right)\) A1
\(\left( { \Rightarrow 2\left( {16 + \frac{{2n - 1}}{4}} \right) = \frac{3}{2}\left( {16 + \frac{{3n - 1}}{4}} \right)} \right)\)
(gdc or algebraic solution) (M1)
n = 63 A2
[9 marks]
\({({a_1} - {a_2})^2} + {({a_2} - {a_3})^2} + {({a_3} - {a_4})^2} + \ldots \)
\( = {({a_1} - {a_1}r)^2} + {({a_1}r - {a_1}{r^2})^2} + ({a_1}{r^2} - {a_1}{r^{^3}}) + \ldots \) M1A1
\( = {\left[ {{a_1}(1 - r)} \right]^2} + {\left[ {{a_1}r(1 - r)} \right]^2} + {\left[ {{a_1}{r^2}(1 - r)} \right]^2} + \ldots + {\left[ {{a_1}{r^{n - 1}}(1 - r)} \right]^2}\) (A1)
Note: This A1 is for the expression for the last term.
\( = a_1^2{(1 - r)^2} + a_1^2{r^2}{(1 - r)^2} + a_1^2{r^4}{(1 - r)^2} + \ldots + a_1^2{r^{2n - 2}}{(1 - r)^2}\) A1
\( = a_1^2{(1 - r)^2}(1 + {r^2} + {r^4} + \ldots + {r^{2n - 2}})\) A1
\( = a_1^2{(1 - r)^2}\left( {\frac{{1 - {r^{2n}}}}{{1 - {r^2}}}} \right)\) M1A1
\( = \frac{{a_1^2(1 - r)(1 - {r^{2n}})}}{{1 + r}}\) AG
[7 marks]
Examiners report
Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between \({S_{2n}}\) and \({S_{3n}}\), and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification.
Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between \({S_{2n}}\) and \({S_{3n}}\), and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification.