Date | November 2010 | Marks available | 6 | Reference code | 10N.1.hl.TZ0.5 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
The mean of the first ten terms of an arithmetic sequence is 6. The mean of the first twenty terms of the arithmetic sequence is 16. Find the value of the 15th term of the sequence.
Markscheme
METHOD 1
5(2a+9d)=60 (or 2a+9d=12) M1A1
10(2a+19d)=320 (or 2a+19d=32) A1
solve simultaneously to obtain M1
a=−3, d=2 A1
the 15th term is −3+14×2=25 A1
Note: FT the final A1 on the values found in the penultimate line.
METHOD 2
with an AP the mean of an even number of consecutive terms equals the mean of the middle terms (M1)
a10+a112=16(or a10+a11=32) A1
a5+a62=6(or a5+a6=12) A1
a10−a5+a11−a6=20 M1
5d+5d=20
d=2 and a=−3(or a5=5 or a10=15) A1
the 15th term is −3+14×2=25(or 5+10×2=25 or 15+5×2=25) A1
Note: FT the final A1 on the values found in the penultimate line.
[6 marks]
Examiners report
Many candidates had difficulties with this question with the given information often translated into incorrect equations.