Date | November 2011 | Marks available | 7 | Reference code | 11N.2.hl.TZ0.12 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 12 | Adapted from | N/A |
Question
In an arithmetic sequence the first term is 8 and the common difference is 14. If the sum of the first 2n terms is equal to the sum of the next n terms, find n.
If a1, a2, a3, … are terms of a geometric sequence with common ratio r≠1, show that (a1−a2)2+(a2−a3)2+(a3−a4)2+…+(an−an+1)2=a21(1−r)(1−r2n)1+r.
Markscheme
S2n=2n2(2(8)+(2n−1)14) (M1)
=n(16+2n−14) A1
S3n=3n2(2×8+(3n−1)14) (M1)
=3n2(16+3n−14) A1
S2n=S3n−S2n⇒2S2n=S3n M1
solve 2S2n=S3n
⇒2n(16+2n−14)=3n2(16+3n−14) A1
(⇒2(16+2n−14)=32(16+3n−14))
(gdc or algebraic solution) (M1)
n = 63 A2
[9 marks]
(a1−a2)2+(a2−a3)2+(a3−a4)2+…
=(a1−a1r)2+(a1r−a1r2)2+(a1r2−a1r3)+… M1A1
=[a1(1−r)]2+[a1r(1−r)]2+[a1r2(1−r)]2+…+[a1rn−1(1−r)]2 (A1)
Note: This A1 is for the expression for the last term.
=a21(1−r)2+a21r2(1−r)2+a21r4(1−r)2+…+a21r2n−2(1−r)2 A1
=a21(1−r)2(1+r2+r4+…+r2n−2) A1
=a21(1−r)2(1−r2n1−r2) M1A1
=a21(1−r)(1−r2n)1+r AG
[7 marks]
Examiners report
Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between S2n and S3n, and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification.
Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between S2n and S3n, and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification.