Date | May 2017 | Marks available | 1 | Reference code | 17M.1.AHL.TZ2.H_3 |
Level | Additional Higher Level | Paper | Paper 1 | Time zone | Time zone 2 |
Command term | Find | Question number | H_3 | Adapted from | N/A |
Question
The 1st, 4th and 8th terms of an arithmetic sequence, with common difference dd, d≠0d≠0, are the first three terms of a geometric sequence, with common ratio rr. Given that the 1st term of both sequences is 9 find
the value of dd;
the value of rr;
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
EITHER
the first three terms of the geometric sequence are 99, 9r9r and 9r29r2 (M1)
9+3d=9r(⇒3+d=3r)9+3d=9r(⇒3+d=3r) and 9+7d=9r29+7d=9r2 (A1)
attempt to solve simultaneously (M1)
9+7d=9(3+d3)29+7d=9(3+d3)2
OR
the 1st1st, 4th4th and 8th8th terms of the arithmetic sequence are
9, 9+3d, 9+7d9, 9+3d, 9+7d (M1)
9+7d9+3d=9+3d99+7d9+3d=9+3d9 (A1)
attempt to solve (M1)
THEN
d=1d=1 A1
[4 marks]
r=43r=43 A1
Note: Accept answers where a candidate obtains dd by finding rr first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in rr.
[1 mark]