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.
Find u1, the first term of the sequence.
Calculate the sum of the first 10 terms of the sequence.
Markscheme
\(\frac{{101.25}}{{135}}\) (M1)
\( = \frac{3}{4}(0.75)\) (A1) (C2)
\({u_1}{\left( {\frac{3}{4}} \right)^4} = 101.25\) (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).
\({S_{10}} = \frac{{320\left( {1 - {{\left( {\frac{3}{4}} \right)}^{10}}} \right)}}{{1 - \left( {\frac{3}{4}} \right)}}\) (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).
Examiners report
The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.
The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.
The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.