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 ${({a_1} - {a_2})^2} + {({a_2} - {a_3})^2} + {({a_3} - {a_4})^2} +  \ldots  + {({a_n} - {a_{n + 1}})^2} = \frac{{a_1^2(1 - r)(1 - {r^{2n}})}}{{1 + r}}$.

${S_{2n}} = \frac{{2n}}{2}\left( {2(8) + (2n - 1)\frac{1}{4}} \right)$     (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]

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.