Date | November 2016 | Marks available | 2 | Reference code | 16N.1.sl.TZ0.9 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
The first two terms of an infinite geometric sequence, in order, are
2log2x, log2x, where x>0.
The first three terms of an arithmetic sequence, in order, are
log2x, log2(x2), log2(x4), where x>0.
Let S12 be the sum of the first 12 terms of the arithmetic sequence.
Find r.
Show that the sum of the infinite sequence is 4log2x.
Find d, giving your answer as an integer.
Show that S12=12log2x−66.
Given that S12 is equal to half the sum of the infinite geometric sequence, find x, giving your answer in the form 2p, where p∈Q.
Markscheme
evidence of dividing terms (in any order) (M1)
egμ2μ1, 2log2xlog2x
r=12 A1 N2
[2 marks]
correct substitution (A1)
eg2log2x1−12
correct working A1
eg2log2x12
S∞=4log2x AG N0
[2 marks]
evidence of subtracting two terms (in any order) (M1)
egu3−u2, log2x−log2x2
correct application of the properties of logs (A1)
eglog2(x2x), log2(x2×1x), (log2x−log22)−log2x
correct working (A1)
eglog212, −log22
d=−1 A1 N3
[4 marks]
correct substitution into the formula for the sum of an arithmetic sequence (A1)
eg122(2log2x+(12−1)(−1))
correct working A1
eg6(2log2x−11), 122(2log2x−11)
12log2x−66 AG N0
[2 marks]
correct equation (A1)
eg12log2x−66=2log2x
correct working (A1)
eg10log2x=66, log2x=6.6, 266=x10, log2(x12x2)=66
x=26.6 (accept p=6610) A1 N2
[3 marks]