Date | November 2020 | Marks available | 2 | Reference code | 20N.2.SL.TZ0.S_6 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_6 | Adapted from | N/A |
Question
An infinite geometric series has first term u1=au1=a and second term u2=14a2-3au2=14a2−3a, where a>0a>0.
Find the common ratio in terms of aa.
Find the values of aa for which the sum to infinity of the series exists.
Find the value of aa when S∞=76S∞=76.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
evidence of dividing terms (in any order) (M1)
eg u1u2, 14a2-3aau1u2, 14a2−3aa
r=14a-3r=14a−3 A1 N2
[2 marks]
recognizing | r |<1|r|<1 (must be in terms of aa) (M1)
eg |14a-3|<1, -1≤14a-3≤1, -4<a-12<4∣∣14a−3∣∣<1, −1≤14a−3≤1, −4<a−12<4
8<a<168<a<16 A2 N3
[3 marks]
correct equation (A1)
eg a1-(14a-3)=76 , a=76(4-14a)a1−(14a−3)=76, a=76(4−14a)
a=765 (=15.2)a=765 (=15.2) (exact) A2 N3
[3 marks]