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.