Date | November 2019 | Marks available | 3 | Reference code | 19N.2.SL.TZ0.S_5 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_5 | Adapted from | N/A |
Question
The first two terms of a geometric sequence are u1=2.1 and u2=2.226.
Find the value of r.
Find the value of u10.
Find the least value of n such that Sn>5543.
Markscheme
valid approach (M1)
eg u1u2, 2.2262.1, 2.226=2.1r
r=1.06 (exact) A1 N2
[2 marks]
correct substitution (A1)
eg 2.1×1.069
3.54790 A1 N2
u10=3.55
[2 marks]
correct substitution into Sn formula (A1)
eg 2.1(1.06n−1)1.06−1, 2.1(1.06n−1)1.06−1>5543, 2.1(1.06n−1)=332.58, sketch of Sn and y=5543
correct inequality for n or crossover values A1
eg n>87.0316, S87=5532.73 and S88=5866.79
n=88 A1 N2
[3 marks]