Show that $d = \frac{a}{2}$.

The seventh term of the arithmetic sequence is $3$. The sum of the first $n$ terms in the arithmetic sequence exceeds the sum of the first $n$ terms in the geometric sequence by at least $200$.

Find the least value of $n$ for which this occurs.

using $r = \frac{{{u_2}}}{{{u_1}}} = \frac{{{u_3}}}{{{u_2}}}$ to form $\frac{{a + 2d}}{{a + 6d}} = \frac{a}{{a + 2d}}$     (M1)

$a(a + 6d) = {(a + 2d)^2}$     A1

$2d(2d - a) = 0\;\;\;$(or equivalent)     A1

since $d \ne 0 \Rightarrow d = \frac{a}{2}$     AG

[3 marks]

substituting $d = \frac{a}{2}$ into $a + 6d = 3$ and solving for $a$ and $d$     (M1)

$a = \frac{3}{4}$ and $d = \frac{3}{8}$     (A1)

$r = \frac{1}{2}$     A1

$\frac{n}{2}\left( {2 \times \frac{3}{4} + (n - 1)\frac{3}{8}} \right) - \frac{{3\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} \ge 200$     (A1)

attempting to solve for $n$     (M1)

$n \ge 31.68 \ldots$

so the least value of $n$ is 32     A1

[6 marks]

Total [9 marks]

Part (a) was reasonably well done. A number of candidates used $r = \frac{{{u_1}}}{{{u_2}}} = \frac{{{u_2}}}{{{u_3}}}$ rather than $r = \frac{{{u_2}}}{{{u_1}}} = \frac{{{u_3}}}{{{u_2}}}$. This invariably led to candidates obtaining $r = 2$ in part (b).

In part (b), most candidates were able to correctly find the first term and the common difference for the arithmetic sequence. However a number of candidates either obtained $r = 2$ via means described in part (a) or confused the two sequences and used ${u_1} = \frac{3}{4}$ for the geometric sequence.