Date | May 2015 | Marks available | 2 | Reference code | 15M.2.sl.TZ2.3 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Write down | Question number | 3 | Adapted from | N/A |
Question
The sum of the first n terms of an arithmetic sequence is given by Sn=6n+n2.
Write down the value of
(i) S1;
(ii) S2.
The nth term of the arithmetic sequence is given by un.
Show that u2=9.
The nth term of the arithmetic sequence is given by un.
Find the common difference of the sequence.
The nth term of the arithmetic sequence is given by un.
Find u10.
The nth term of the arithmetic sequence is given by un.
Find the lowest value of n for which un is greater than 1000.
The nth term of the arithmetic sequence is given by un.
There is a value of n for which
u1+u2+…+un=1512.
Find the value of n.
Markscheme
(i) S1=7 (A1)
(ii) S2=16 (A1)
(u2=) 16−7=9 (M1)(AG)
Note: Award (M1) for subtracting 7 from 16. The 9 must be seen.
OR
16−7−7=2
(u2=) 7+(2−1)(2)=9 (M1)(AG)
Note: Award (M1) for subtracting twice 7 from 16 and for correct substitution in correct arithmetic sequence formula.
The 9 must be seen.
Do not accept: 9−7=2, u2=7+(2−1)(2)=9.
u1=7 (A1)(ft)
d=2 (=9−7) (A1)(ft)(G2)
Notes: Follow through from their S1 in part (a)(i).
7+2×(10−1) (M1)
Note: Award (M1) for correct substitution in the correct arithmetic sequence formula. Follow through from their parts (a)(i) and (c).
=25 (A1)(ft)(G2)
Note: Award (A1)(ft) for their correct tenth term.
7+2×(n−1)>1000 (A1)(ft)(M1)
Note: Award (A1)(ft) for their correct expression for the nth term, (M1) for comparing their expression to 1000. Accept an equation. Follow through from their parts (a)(i) and (c).
n=498 (A1)(ft)(G2)
Notes: Answer must be a natural number.
6n+n2=1512ORn2(14+2(n−1))=1512OR
Sn=1512OR7+9+…+un=1512 (M1)
Notes: Award (M1) for equating the sum of the first n terms to 1512. Accept a sum of at least the first 7 correct terms.
n=36 (A1)(G2)
Note: If n=36 is seen without working, award (G2). Award a maximum of (M1)(A0) if −42 is also given as a solution.