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Date	November 2021	Marks available	4	Reference code	21N.2.SL.TZ0.6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	6	Adapted from	N/A

Question

The sum of the first $n$ terms of a geometric sequence is given by $S_{n} = Σ_{r = 1}^{n} \frac{2}{3} {(\frac{7}{8})}^{r}$ .

Find the first term of the sequence, $u_{1}$ .

[2]

Find $S_{\infty}$ .

[3]

Find the least value of $n$ such that $S_{\infty} - S_{n} < 0.001$ .

[4]

Markscheme

$u_{1} = S_{1} = \frac{2}{3} \times \frac{7}{8}$ (M1)

$= \frac{14}{24} (= \frac{7}{12} = 0.583333 \dots)$ A1

[2 marks]

$r = \frac{7}{8} (= 0.875)$ (A1)

substituting their values for $u_{1}$ and $r$ into $S_{\infty} = \frac{u_{1}}{1 - r}$ (M1)

$= \frac{14}{3} (= 4.66666 \dots)$ A1

[3 marks]

attempt to substitute their values into the inequality or formula for $S_{n}$ (M1)

$\frac{14}{3} - Σ_{r = 1}^{n} \frac{2}{3} {(\frac{7}{8})}^{r} < 0.001$ OR $S_{n} = \frac{\frac{7}{12} (1 - {(\frac{7}{8})}^{n})}{(1 - \frac{7}{8})}$

attempt to solve their inequality using a table, graph or logarithms

(must be exponential) (M1)

Note: Award (M0) if the candidate attempts to solve $S_{\infty} - u_{n} < 0.001$ .

correct critical value or at least one correct crossover value (A1)

$63.2675 \dots$ OR $S_{\infty} - S_{63} = 0.001036 \dots$ OR $S_{\infty} - S_{64} = 0.000906 \dots$

OR $S_{\infty} - S_{63} - 0.001 = 0.0000363683 \dots$ OR $S_{\infty} - S_{64} - 0.001 = 0.0000931777 \dots$

least value is $n = 64$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series

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Topic 1—Number and algebra » SL 1.8—Sum of infinite geo sequence

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