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Date November 2019 Marks available 2 Reference code 19N.1.SL.TZ0.T_7
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number T_7 Adapted from N/A

Question

A geometric sequence has a first term of  8 3  and a fourth term of 9 .

Find the common ratio.

[2]
a.

Write down the second term of this sequence.

[1]
b.

The sum of the first  k  terms is greater than  2500 .

Find the smallest possible value of k .

[3]
c.

Markscheme

9 = ( 8 3 ) r 3       (M1)

Note: Award (M1) for correctly substituted geometric sequence formula equated to 9 .

( r = ) 1.5 ( 3 2 )          (A1) (C2)

[2 marks]

a.

4          (A1)(ft)  (C1)

Note: Follow through from part (a).

[1 mark]

b.

2500 < ( 8 3 ) ( ( 1.5 ) k 1 ) 1.5 1        (M1)

Note: Award (M1) for their correctly substituted geometric series formula compared to 2500 .

k = 15.2 ( 15.17319 )         (A1)(ft)

( k = ) 16         (A1)(ft)    (C3)

Note: Answer must be an integer for the final (A1)(ft) to be awarded.
Follow through from part (a).

[3 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

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

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