Arithmetic Series [64 marks]

In an arithmetic sequence, u2 = 5 and u3 = 11.

1a. Find the common difference. [2 marks]

Markscheme
valid approach (M1)
eg 11 − 5, 11 = 5 + d
d=6 A1 N2
[2 marks]

1b. Find the first term. [2 marks]

Markscheme
valid approach (M1)
eg u2 − d, 5 − 6, u1 + (3 − 1) (6) = 11
u1 = −1 A1 N2
[2 marks]

1c. Find the sum of the first 20 terms. [2 marks]

Markscheme
correct substitution into sum formula
20 20
eg 2 (2 (−1) + 19 (6)), 2 (−1 + 113) (A1)
S20 = 1120 A1 N2
[2 marks]
 In an arithmetic sequence, u1 = −5 and d = 3.

2. Find u8. [2 marks]

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
correct working (A1)
eg −5 + (8 − 1)(3)
u8 = 16 A1 N2

[2 marks]

Sergei is training to be a weightlifter. Each day he trains at the local gym by lifting
a metal bar that has heavy weights attached. He carries out successive lifts. After
each lift, the same amount of weight is added to the bar to increase the weight to
be lifted.
The weights of each of Sergei’s lifts form an arithmetic sequence.
Sergei’s friend, Yuri, records the weight of each lift. Unfortunately, last Monday,
Yuri misplaced all but two of the recordings of Sergei’s lifts.
On that day, Sergei lifted 21 kg on the third lift and 46 kg on the eighth lift.

3a. For that day find how much weight was added after each lift. [2 marks]

Markscheme
5d = 46 − 21 OR u1 + 2d = 21 and u1 + 7d = 46 (M1)
Note: Award (M1) for a correct equation in d or for two correct equations in
u1 and d.
(d =) 5 (kg) (A1) (C2)
[2 marks]

3b. For that day find the weight of Sergei’s first lift. [2 marks]
 Markscheme
u1 + 2 × 5 = 21 (M1)
OR
u1 + 7 × 5 = 46 (M1)
Note: Award (M1) for substitution of their d into either of the two equations.
(u1 =) 11 (kg) (A1)(ft) (C2)
Note: Follow through from part (a)(i).
[2 marks]

3c. On that day, Sergei made 12 successive lifts. Find the total combined [2 marks]
weight of these lifts.

Markscheme
12
2 (2 × 11 + (12 − 1) × 5) (M1)
Note: Award (M1) for correct substitution into arithmetic series formula.
= 462 (kg) (A1)(ft) (C2)
Note: Follow through from parts (a) and (b).
[2 marks]

A new café opened and during the first week their profit was $60.
The café’s profit increases by $10 every week.

4. Calculate the café’s total profit for the first 12 weeks. [3 marks]
 Markscheme
12
2 (2 × 60 + 11 × 10) (M1)(A1)(ft)
Note: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for
correct substitution. Follow through from their first term and common
difference in part (a).
= ($) 1380 (A1)(ft)(G2)
[3 marks]

The company Snakezen’s Ladders makes ladders of different lengths. All the
ladders that the company makes have the same design such that:
the first rung is 30 cm from the base of the ladder,
the second rung is 57 cm from the base of the ladder,
the distance between the first and second rung is equal to the distance between
all adjacent rungs on the ladder.
The ladder in the diagram was made by this company and has eleven equally
spaced rungs.

5a. Find the distance from the base of this ladder to the top rung. [3 marks]
 Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
30 + (11 − 1) × 27 (M1)(A1)

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

= 300 (cm) (A1) (C3)

Note: Units are not required.

[3 marks]

5b. The company also makes a ladder that is 1050 cm long. [3 marks]
Find the maximum number of rungs in this 1050 cm long ladder.

Markscheme
1050 ⩾ 30 + (n − 1) × 27 (M1)(A1)(ft)

Note: Award (M1) for substituted arithmetic sequence formula ⩽ 1050,

accept an equation, (A1) for correct substitutions.

n = 38 (A1)(ft) (C3)

Note: Follow through from their 27 in part (a). The answer must be an
integer and rounded down.

[3 marks]
 Tomás is playing with sticks and he forms the first three diagrams of a pattern.
These diagrams are shown below.

Tomás continues forming diagrams following this pattern.

6a. Diagram n is formed with 52 sticks. Find the value of n. [3 marks]

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
4 + 3(n − 1) = 52 (M1)(A1)

Note: Award (M1) for substitution into the formula of the nth term of an
arithmetic sequence, (A1) for correct substitution.

n = 17 (A1) (C3)
[3 marks]

Tomás forms a total of 24 diagrams.

6b. Find the total number of sticks used by Tomás for all 24 diagrams. [3 marks]
 Markscheme
24
2 (2 × 4 + 23 × 3)OR 24
2 (4 + 73) (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sum of the first n terms of an
arithmetic sequence formula, (A1)(ft) for their correct substitution, consistent
with part (a).

924 (A1)(ft) (C3)

Note: Follow through from part (a).

[3 marks]

7a. One of the locations in the 2016 Olympic Games is an amphitheatre. The [2 marks]
number of seats in the first row of the amphitheatre, u1 , is 240. The
number of seats in each subsequent row forms an arithmetic sequence. The
number of seats in the sixth row, u6 , is 270.
Calculate the value of the common difference, d.

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
270 = 240 + d (6 − 1) (M1)
OR
270−240
d= 5
(M1)
Note: Award (M1) for correct substitution into the arithmetic sequence
formula.
(d =) 6 (A1) (C2)

7b. There are 20 rows in the amphitheatre. [2 marks]

Find the total number of seats in the amphitheatre.
 Markscheme
20
2 [2 × 240 + 19 × their d] (M1)
Note: Award (M1) for correct substitution into sum of an arithmetic sequence.
OR
u20 = 354
20
S20 = 2 [240 + 354] (M1)
Note: Award (M1) for correct substitution into sum of an arithmetic sequence.
OR
adding20 terms consistent with their d (M1)
= 5940 (A1)(ft) (C2)
Note: Follow through from (a).

7c. Anisha visits the amphitheatre. She estimates that the amphitheatre has [2 marks]
6500 seats.
Calculate the percentage error in Anisha’s estimate.

Markscheme
∣∣ 6500−5940 ∣∣ × 100 (M1)
5940
Note: Award (M1) for correct substitution into percentage error formula.
= 9.43 (%) (9.42760...) (A1)(ft) (C2)
Note: Follow through from (b).

8. The fifth term of an arithmetic sequence is equal to 6 and the sum of the [6 marks]
first 12 terms is 45.
Find the first term and the common difference.
 Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
use of either un = u1 + (n − 1)d or Sn = n (2u
2 1 + (n − 1)d) M1
u1 + 4d = 6 (A1)
12
2 (2u1 + 11d) = 45 (A1)
⇒ 4u1 + 22d = 15
attempt to solve simultaneous equations M1
4(6 − 4d) + 22d = 15
6d = −9 ⇒ d = −1.5 A1
u1 = 12 A1
[6 marks]

The sum of the first n terms of an arithmetic sequence is given by Sn = 6n + n2 .

9a. Write down the value of [2 marks]

(i) S1 ;
(ii) S2 .

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(i) S1 = 7 (A1)
(ii) S2 = 16 (A1)

nth term of the arithmetic sequence is given by un .

9b. The [1 mark]
Show that u2 = 9.
 Markscheme
(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.

9c. The nth term of the arithmetic sequence is given by un . [2 marks]

Find the common difference of the sequence.

Markscheme
u1 = 7 (A1)(ft)
d = 2 (= 9 − 7) (A1)(ft)(G2)
Notes: Follow through from their S1 in part (a)(i).

9d. Thenth term of the arithmetic sequence is given by un . [2 marks]

Find u10 .

Markscheme
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.
9e. Thenth term of the arithmetic sequence is given by un . [3 marks]
Find the lowest value of n for which un is greater than 1000.

Markscheme
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.

9f. Thenth term of the arithmetic sequence is given by un . [2 marks]

There is a value of n for which
u1 + u2 + … + un = 1512.
Find the value of n.

Markscheme
6n + n2 = 1512 OR n (14 +
2 2(n − 1)) = 1512 OR
Sn = 1512 OR 7 + 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.
 The number of apartments in a housing development has been increasing by a
constant amount every year.
At the end of the first year the number of apartments was 150, and at the end of
the sixth year the number of apartments was 600.
The number of apartments, y, can be determined by the equation y = mt + n,
where t is the time, in years.

10a. Find the value of m . [2 marks]

Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
600−150
6−1
(M1)

OR
600 = 150 + (6 − 1)m (M1)
Note: Award (M1) for correct substitution into gradient formula or arithmetic
sequence formula.

= 90 (A1) (C2)

10b. Find the value of n. [2 marks]

Markscheme
150 = 90 × (1) + n (M1)
Note: Award (M1) for correct substitution of their gradient and one of the
given points into the equation of a straight line.

n = 60 (A1)(ft) (C2)
Note: Follow through from part (a).

In an arithmetic sequence, the third term is 10 and the fifth term is 16.

11a. Find the common difference. [2 marks]

 Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
attempt to find d (M1)
16−10
eg 2 , 10 − 2d = 16 − 4d, 2d = 6, d = 6
d=3 A1 N2
[2 marks]

11b. Find the first term. [2 marks]

Markscheme
correct approach (A1)
eg 10 = u1 + 2 × 3, 10 − 3 − 3
u1 = 4 A1 N2
[2 marks]

11c. Find the sum of the first 20 terms of the sequence. [3 marks]

Markscheme
correct substitution into sum or term formula (A1)
20
eg 2 (2 × 4 + 19 × 3), u20 = 4 + 19 × 3
correct simplification (A1)
eg 8 + 57, 4 + 61
S20 = 650 A1 N2
[3 marks]

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