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Oxford-Math Analysis and Approaches SL Question Booklet 2025

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Oxford-Math Analysis and Approaches SL Question Booklet 2025

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End of chapter test




1 From patterns to generalizations:
sequences and series
Section A. A calculator is not allowed

1 The nth term of an arithmetic sequence is given by un= 4 + 3n

a Write down the first four terms.

b Find the common difference.

c Find the value of n if the nth term is 109.

d Find the sum of the first 21 terms.

2 In an arithmetic sequence, the 40th term is 144 and the sum of the first 40 terms is 2640. Find

a the first term

b the common difference.

2 2 2
3 Given the sequence ,   ,   ,  2 :
27 9 3

a Find the common ratio.

b Find the eighth term.

4 A school theatre has 24 rows of seats. There are 18 seats in the first row and each subsequent
row had two more seats than the previous row. What is the seating capacity of the theatre?

5 Find the first two terms of an arithmetic sequence where the sixth term is 21 and the sum of
the first 17 terms is zero.

6 a How many terms are there in the expansion of ( a + b ) ?
10




b Show the full expansions of (3x + 2y ) .
4




7 Evaluate
5 6
a ∑ (x
n =1
2
)
+ 2    b ∑2
i =2
i




Section B. A calculator is allowed

8 A geometric sequence has a 2nd term of 6 and a 5th term of 162.

a Find the common ratio.

b Find the 10th term.

c Find the sum of the first 8 terms.

9 A rose bush is 1.67m tall when planted, and each week its height increases by 4%. How tall
will it be after 10 weeks?



© Oxford University Press 2019 End of chapter test 1

, End of chapter test




10 The Fibonacci sequence is named after Italian mathematician Leonardo of Pisa, who was known as
Fibonacci.

In this Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, …,

a Find the 10th term of the Fibonacci sequence.

b Write a recursive formula for the Fibonacci sequence.

11 a Explain the condition for a geometric series to be convergent.

10 10
Given the series 10 − + − …,
3 9

b Find the common ratio

c Find the sum to infinity.
9
 2 
12 Find the constant term in the expansion of  x − 2 
 x 

13 Yosef drops a basketball from his bedroom window, which is 3m off the ground. After each bounce,
the basketball comes back to 75% of its previous height. If it keeps on bouncing forever, what
vertical distance, to the nearest metre, will it travel?

14 Tafari starts a job and deposits $500 from the first salary into a bank account producing 4%
interest every month. Each month after that, Tafari deposits an additional $100.

a Calculate the amount Tafari has in the account at the end of each month for the first three
months.

b Write a recursive formula for the amount of money in the account.

c Show that the formula for the amount of money in the account for year n is

(
500 (1.04 ) + 2500 (1.04 ) − 1
n n
)
d Use the formula to find the amount of money in Tafani’s account after 24 months.




© Oxford University Press 2019 2

, End of chapter test




Answers

1 a 7, 10, 13, 16 n
d=
Sn
2
(u1 + un )
b d = 10 − 3 = 7
u1 = 7, n = 21, u21 = 4 + 3(21) = 67
c 4 + 3n =
109
21
3n = 105 S21 =
2
(7 + 67) = 777
n = 35

n b un = u1 + ( n − 1) d
2 a=
Sn
2
(u1 + un )
144 =−12 + ( 40 − 1) d
40
2640
=
2
(u1 + 144)
39d = 156

132
= (u1 + 144) d =4

u1 = −12

2 2 2
3 Given the sequence ,   ,   ,  2
27 9 3

2
9  2  7 2 7
a=r = 3 b u8 = 4
 3 = 3  3 =2 × 3 =162
2  27  3 
27

n
4 sn
=
2
(
2u1 + ( n − 1) d )
24
s24
=
2
(
2 (18 ) + (24 − = )
1) 2 984 seats


5 un = u1 + ( n − 1) d 21
= u1 + 5d
n 17
sn
=
2
( ) d 0 2 (2u1 + 16d )
2u1 + ( n − 1= )
This gives the simultaneous equations

u1 + 5d =
21

u1 + 8d =
0

Solve to find d = –7 and u1 = 56.

The first two terms are 56 and 49.

6 a 11

b Using the 4th row of Pascal’s triangle, 1 4 6 4 1,

1 (3x ) + 4 (3x ) (2y ) + 6 (3x ) (2y ) + 4 (3x ) (2y ) + 1 (2y )
4 3 2 2 1 3 4




=81x 4 + 216 x 3y + 216 x 2y 2 + 96 xy 3 + 16y 4

7 a 3 + 6 + 11 + 18 + 27  65
= b 4 +8 + 16 + 32 + 64 =124




© Oxford University Press 2019 3

, End of chapter test




8 a 6r3 = 162

r3 = 27

r=3

b = ( u1 ) r n −1
un

u10 = 2 × 39 = 39366


c S8
=
(
2 38 − 1
= 6560
)
3 −1

9 = ( u1 ) r n −1
un

= (=
u10   1.67 ) (1.04)10−1 2.38m
10 a 55

b un +=
1
un + un −1

11 a −1 < r < 1

10 10
Given the series 10 − + −…
3 9

10 1
b r =
− ÷ 10 =

3 3

10 10
c S∞
= = = 7.5
 1 4
1 − −  3
 3

3
9  −2 
12   x 6  2  = −672
3
  x 

13




u1 = 3, r = 0.75

3 3
S∞
= = = 12
1 − 0.75 0.25

To account for upward and downward movement, multiply by 2, but the original drop of 3 only
occurred once, so subtract 3 from the answer.

Vertical distance = (12 × 2) − 3= 21metres


© Oxford University Press 2019 4

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