| Date | November 2015 | Marks available | 2 | Reference code | 15N.2.sl.TZ0.4 |
| Level | SL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The first three terms of a geometric sequence are u1=0.64, u2=1.6, and u3=4.
Find the value of r.
[2]
a.
Find the value of S6.
[2]
b.
Find the least value of n such that Sn>75000.
[3]
c.
Markscheme
valid approach (M1)
egu1u2, 41.6, 1.6=r(0.64)
r=2.5(=52) A1 N2
[2 marks]
a.
correct substitution into S6 (A1)
eg0.64(2.56−1)2.5−1
S6=103.74 (exact), 104 A1 N2
[2 marks]
b.
METHOD 1 (analytic)
valid approach (M1)
eg0.64(2.5n−1)2.5−1>75000, 0.64(2.5n−1)2.5−1=75000
correct inequality (accept equation) (A1)
egn>13.1803, n=13.2
n=14 A1 N1
METHOD 2 (table of values)
both crossover values A2
egS13=63577.8, S14=158945
n=14 A1 N1
[3 marks]
Total [7 marks]
c.
Examiners report
[N/A]
a.
[N/A]
b.
[N/A]
c.
Syllabus sections
Topic 1 - Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
Show 92 related questions
