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Date November 2011 Marks available 3 Reference code 11N.1.sl.TZ0.12
Level SL only Paper 1 Time zone TZ0
Command term Calculate Question number 12 Adapted from N/A

Question

Consider the sequence

\[{\text{512, 128, 32, 8, }} \ldots \]

Calculate the exact value of the ninth term of the sequence.

[3]
a.

Calculate the least number of terms required for the sum of the sequence to be greater than 682.6

[3]
b.

Markscheme

\({u_9} = 512{\left( {\frac{1}{4}} \right)^8}\)     (M1)(A1)

Notes: Award (M1) for substituted geometric sequence formula, (A1) for correct substitution.

 

OR

If a list is used, award (M1) for a list of 9 terms, (A1) for all 9 terms correct.     (M1)(A1)

\( = \frac{1}{{128}}\) (\(0.0078125\))     (A1)     (C3)

Note: Award (A1) for exact answer only.

[3 marks]

a.

\(\frac{{512\left( {1 - {{\left( {\frac{1}{4}} \right)}^n}} \right)}}{{1 - \left( {\frac{1}{4}} \right)}} > 682.6\)     (M1)(A1)(ft)

Notes: Award (M1) for setting substituted geometric sum formula \( > 682.6\) (A1)(ft) for correct substitution into geometric sum formula. Follow through from their common ratio.

 

OR

If list is used, award (M1) for S(6) and S(7) seen, values don’t have to be correct.

(A1) for correct S(6) and S(7). (S(6) \( = 682.5\) and S(7) \( = 682.625\)).     (M1)(A1)

\(n = 7\)     (A1)(ft)     (C3)

Notes: Follow through from their common ratio. Do not award the final (A1)(ft) if \(n\) is less than \(5\) or if \(n\) is not an integer.

[3 marks]

b.

Examiners report

The upper quartile of candidates scored well on this question with the vast majority scoring more than 4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A fundamental error in part (a) resulted in many less able candidates using a common ratio of \(4\) instead of \({\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}\)
. Where lists were used in either part of the question, they were either invariably incomplete or contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign was allowed) did earn many candidates two marks here. The required answer of \(7\) however did not always follow.

a.

The upper quartile of candidates scored well on this question with the vast majority scoring more than 4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A fundamental error in part (a) resulted in many less able candidates using a common ratio of \(4\) instead of \({\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}\)
. Where lists were used in either part of the question, they were either invariably incomplete or contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign was allowed) did earn many candidates two marks here. The required answer of \(7\) however did not always follow.

b.

Syllabus sections

Topic 1 - Number and algebra » 1.8 » Geometric sequences and series.
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