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Date May 2022 Marks available 1 Reference code 22M.1.SL.TZ1.8
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Write down Question number 8 Adapted from N/A

Question

Consider the series lnx+plnx+13lnx+, where x, x>1 and p, p0.

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference d.

Show that p=±13.

[2]
a.i.

Given that p>0 and S=3+3, find the value of x.

[3]
a.ii.

Show that p=23.

[3]
b.i.

Write down d in the form klnx, where k.

[1]
b.ii.

The sum of the first n terms of the series is -3lnx.

Find the value of n.

[6]
b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms        M1

plnxlnx=13lnxplnx  OR  13lnx=lnxr2  OR  plnx=lnx13p

 

Note: Candidates may use lnx1+lnxp+lnx13 and consider the powers of x in geometric sequence

Award M1 for p1=13p.


OR

r=p  and  r2=13        M1


THEN

p2=13  OR  r=±13          A1

p=±13          AG

 

Note: Award M0A0 for r2=13 or p2=13 with no other working seen.

 

[2 marks]

a.i.

lnx1-13  =3+3           (A1)

lnx=3-33+3-33  OR  lnx=3-3+3-1  lnx=2          A1

x=e2          A1

 

[3 marks]

a.ii.

METHOD 1

attempt to find a difference from consecutive terms or from u2          M1

correct equation          A1

plnx-lnx=13lnx-plnx  OR  13lnx=lnx+2plnx-lnx


Note:
Candidates may use lnx1+lnxp+lnx13+ and consider the powers of x in arithmetic sequence.

Award M1A1 for p-1=13-p

 

2plnx=43lnx  2p=43          A1

p=23          AG

 

METHOD 2

attempt to use arithmetic mean u2=u1+u32          M1

plnx=lnx+13lnx2          A1

2plnx=43lnx  2p=43          A1

p=23          AG

 

METHOD 3

attempt to find difference using u3          M1

13lnx=lnx+2d  d=-13lnx

 

u2=lnx+1213lnx-lnx  OR  plnx-lnx=-13lnx          A1

plnx=23lnx          A1

p=23          AG

 

[3 marks]

b.i.

d=-13lnx       A1

 

[1 mark]

b.ii.

METHOD 1

Sn=n22lnx+n-1×-13lnx

attempt to substitute into Sn and equate to -3lnx           (M1)

n22lnx+n-1×-13lnx=-3lnx

correct working with Sn (seen anywhere)           (A1)

n22lnx-n3lnx+13lnx  OR  nlnx-nn-16lnx  OR  n2lnx+4-n3lnx

correct equation without lnx          A1

n273-n3=-3  OR  n-nn-16=-3 or equivalent


Note:
Award as above if the series 1+p+13+ is considered leading to n273-n3=-3.


attempt to form a quadratic =0           (M1)

n2-7n-18=0

attempt to solve their quadratic           (M1)

n-9n+2=0

n=9          A1

 

METHOD 2

listing the first 7 terms of the sequence           (A1)

lnx+23lnx+13lnx+0-13lnx-23lnx-lnx+

recognizing first 7 terms sum to 0           M1

8th term is -43lnx           (A1)

9th term is -53lnx           (A1)

sum of 8th and 9th term =-3lnx           (A1)

n=9          A1

 

[6 marks]

b.iii.

Examiners report

Many candidates were able to identify the key relationship between consecutive terms for both geometric and arithmetic sequences. Substitution into the infinity sum formula was good with solving involving the natural logarithm done quite well. The complexity of the equation formed using 𝑆𝑛 was a stumbling block for some candidates. Those who factored out and cancelled the ln𝑥 expression were typically successful in solving the resulting quadratic.

a.i.
[N/A]
a.ii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
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