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Date May 2013 Marks available 2 Reference code 13M.1.sl.TZ1.7
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 7 Adapted from N/A

Question

The fourth term, u4, of a geometric sequence is 135. The fifth term, u5, is 101.25 .

Find the common ratio of the sequence.

[2]
a.

Find u1, the first term of the sequence.

[2]
b.

Calculate the sum of the first 10 terms of the sequence.

[2]
c.

Markscheme

\(\frac{{101.25}}{{135}}\)     (M1)

\( = \frac{3}{4}(0.75)\)     (A1)     (C2)

a.

\({u_1}{\left( {\frac{3}{4}} \right)^4} = 101.25\)     (M1)

OR

\({u_1}{\left( {\frac{3}{4}} \right)^3} = 135\)     (M1)

OR

(by list)

\({u_3} = 180,{\text{ }}{u_2} = 240\)     (M1)

Notes: Award (M1) for their correct substitution in geometric sequence formula, or stating explicitly \({u_3}\) and \({u_2}\).


\(({u_1} = )320\)     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).

b.

\({S_{10}} = \frac{{320\left( {1 - {{\left( {\frac{3}{4}} \right)}^{10}}} \right)}}{{1 - \left( {\frac{3}{4}} \right)}}\)     (M1)


Notes: Award (M1) for their correct substitution in geometric series formula.

    Accept a list of all their ten geometric terms.


= 1210 (1207.918...)     (A1)(ft)     (C2)


Note: Follow through from their parts (a) and (b).

c.

Examiners report

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

a.

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

b.

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

c.

Syllabus sections

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