Date | May 2018 | Marks available | 6 | Reference code | 18M.1.SL.TZ2.S_7 |
Level | Standard Level | Paper | Paper 1 | Time zone | Time zone 2 |
Command term | Find | Question number | S_7 | Adapted from | N/A |
Question
An arithmetic sequence has u1=logc(p)u1=logc(p) and u2=logc(pq)u2=logc(pq), where c>1c>1 and p,q>0p,q>0.
Let p=c2p=c2 and q=c3q=c3. Find the value of 20∑n=1un20∑n=1un.
Markscheme
METHOD 1 (finding u1u1 and d)
recognizing ∑=S20∑=S20 (seen anywhere) (A1)
attempt to find u1u1 or d using logcck=klogcck=k (M1)
eg logcclogcc, 3logcc3logcc, correct value of u1u1 or d
u1u1 = 2, d = 3 (seen anywhere) (A1)(A1)
correct working (A1)
eg S20=202(2×2+19×3),S20=202(2+59),10(61)S20=202(2×2+19×3),S20=202(2+59),10(61)
20∑n=1un20∑n=1un = 610 A1 N2
METHOD 2 (expressing S in terms of c)
recognizing ∑=S20∑=S20 (seen anywhere) (A1)
correct expression for S in terms of c (A1)
eg 10(2logcc2+19logcc3)10(2logcc2+19logcc3)
logcc2=2,logcc3=3logcc2=2,logcc3=3 (seen anywhere) (A1)(A1)
correct working (A1)
eg S20=202(2×2+19×3),S20=202(2+59),10(61)S20=202(2×2+19×3),S20=202(2+59),10(61)
20∑n=1un20∑n=1un = 610 A1 N2
METHOD 3 (expressing S in terms of c)
recognizing ∑=S20∑=S20 (seen anywhere) (A1)
correct expression for S in terms of c (A1)
eg 10(2logcc2+19logcc3)10(2logcc2+19logcc3)
correct application of log law (A1)
eg 2logcc2=logcc4,19logcc3=logcc57,10(logc(c2)2+logc(c3)19),10(logcc4+logcc57),10(logcc61)2logcc2=logcc4,19logcc3=logcc57,10(logc(c2)2+logc(c3)19),10(logcc4+logcc57),10(logcc61)
correct application of definition of log (A1)
eg logcc61=61,logcc4=4,logcc57=57logcc61=61,logcc4=4,logcc57=57
correct working (A1)
eg S20=202(4+57),10(61)S20=202(4+57),10(61)
20∑n=1un20∑n=1un = 610 A1 N2
[6 marks]