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 83 and a fourth term of 9.
Find the common ratio.
Write down the second term of this sequence.
The sum of the first k terms is greater than 2500.
Find the smallest possible value of k.
Markscheme
9=(83)r3 (M1)
Note: Award (M1) for correctly substituted geometric sequence formula equated to 9.
(r=)1.5(32) (A1) (C2)
[2 marks]
4 (A1)(ft) (C1)
Note: Follow through from part (a).
[1 mark]
2500<(83)((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]