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Date November 2019 Marks available 2 Reference code 19N.2.SL.TZ0.T_3
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find and Hence Question number T_3 Adapted from N/A

Question

Maegan designs a decorative glass face for a new Fine Arts Centre. The glass face is made up of small triangular panes. The first three levels of the glass face are illustrated in the following diagram.

The 1st level, at the bottom of the glass face, has 5 triangular panes. The 2nd level has 7 triangular panes, and the 3rd level has 9 triangular panes. Each additional level has 2 more triangular panes than the level below it.

Maegan has 1000 triangular panes to build the decorative glass face and does not want it to have any incomplete levels.

Find the number of triangular panes in the 12th level.

[3]
a.

Show that the total number of triangular panes, Sn, in the first n levels is given by:

Sn=n2+4n.

[3]
b.

Hence, find the total number of panes in a glass face with 18 levels.

[2]
c.

Find the maximum number of complete levels that Maegan can build.

[3]
d.

Each triangular pane has an area of 1.84m2.

Find the total area of the decorative glass face, if the maximum number of complete levels were built. Express your area to the nearest m2.

[4]
e.

Markscheme

u12=5+12-1×2      (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.

27        (A1)(G3)

[3 marks]

a.

Sn=n22×5+n-12      (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions.

Sn=n28+2n  OR  Sn=n5+n-1        (M1)

Note: Award (M1) for evidence of expansion and simplification, or division by 2 leading to the final answer.

Sn=n2+4n        (AG)

Note: The final line must be seen, with no incorrect working, for the final (M1) to be awarded.

[3 marks]

b.

S18=182+4×18      (M1)

Note: Award (M1) for correctly substituted formula for Sn.

S18= 396        (A1)

Note: The use of “hence” in the question paper means that the Sn formula (from part (b)) must be used.

[2 marks]

c.

1000=n2+4n  OR  1000=n210+n-12 (or equivalent)      (M1)

Note: Award (M1) for equating Sn to 1000 or for equating the correctly substituted sum of arithmetic sequence formula to 1000.

OR

a sketch of the graphs Sn=n2+4n and Sn=1000  intersecting       (M1)

Note: Award (M1) for a sketch of a quadratic and a horizontal line with at least one point of intersection.

OR

a sketch of n2+4n-1000 intersecting the x-axis      (M1)

Note: Award (M1) for a sketch of n2+4n-1000 with at least one x-intercept.

n= 29.6859  OR  -2+2251        (A1)

Note: Award (A1) for 29.6859 or -2+2251 seen. Can be implied by a correct final answer.

n= 29          (A1)(ft)(G2)

Note: Do not accept 30. Award a maximum of (M1)(A1)(A0) if two final answers are given. Follow though from their unrounded answer.

OR

S30=1020  and  S29=957          (A2)

Note: Award (A2) for both “crossover” values seen. Do not split this (A2) mark.

n= 29          (A1)(G2)

 

[3 marks]

d.

A= 292+4×29×1.84      (M1)(M1)

Note: Award (M1) for their correct substitution to find the total number of triangular panes. Award (M1) for multiplying their number of panes by 1.84.

OR

A= 957×1.84          (A1)(ft)(M1)

Note: Award (A1)(ft) for their 957 seen. Award (M1) for multiplying their number of panes by 1.84. Follow through from part (d).

A= 1760.88m2          (A1)(ft)(G2)

A= 1761m2          (A1)(ft)(G3)

[4 marks]

e.

Examiners report

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Syllabus sections

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

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