Material Taken From:
Mathematics
for the international student
Mathematical Studies SL
Mal Coad, Glen Whiffen, John Owen, Robert Haese, Sandra Haese and Mark
Bruce
Haese and Haese Publications, 2004
AND
Mathematical Studies Standard Level
Peter Blythe, Jim Fensom, Jane Forrest and Paula Waldman de Tokman
Oxford University Press, 2012
Applying Arithmetic
Sequences and Series
Susan wants to buy a flat. She has to pay for the flat in 20 year installments. The
first installment is 5500 €. Each installment is 500 € more than the one before.
A. Write down the value of the second and third installment.
B. Calculate the value of the final installment.
C. Show that the total amount Susan would pay for the flat is 205000 €.
u1 = 5500 and d = 500
u2 = 6000 € and u3 = 6500 €
u20 = u1 + (n – 1)d = 5500 + (20 – 1)500 = 15,000 €
n
S n   2u1  (n  1)d 
2
S20 =
𝟐𝟎
𝟐
(2(5500) + (20 – 1)500)
S20 = 205,000 €
The sales of Smartphones are growing every year. At the end of 2006, the
number sold was 25,000,000. At the end of 2010, the number sold was
35,800,000. Assuming that the sales follow an arithmetic sequence, calculate:
A. the number pf Smartphones sold at the end of 2008
B. the predicted number of Smartphones sold at the end of 2015
u1 = 25,000,000
AND
u5 = u1 + 4d = 35,800,000
35,800,000 = 25,000,000 + 4d
d = 2,700,000
2008 would be u3
u3 = u1 + 2d = 25,000,000 + 2(2,700,000) = 30,400,00
2015 would be u10
u10 = u1 + 9d = 25,000,000 + 9(2,700,000) = 49,300,00
Applying Geometric
Sequences and Series
Penelope is starting her first job. She will earn $24,000 in the first year
and her salary will increase by 4% every year. Calculate how much
Penelope will earn in her fourth year of work.
u1 = $24,000
AND
r = 1.04
un  u1r n 1
u4 = 2400 x 1.043
u4 = $26996.74
A ball is vertically dropped. It reaches a height of 1.6 m on the first bounce.
The height of each subsequent bounce is 80% of the previous bounce.
A. Find the height the ball reaches on the 6th bounce.
B. Find the sum of the first seven terms of this sequence.
u1 = 1.6 m
un  u1r n 1
Sn = u1 (r n – 1)
(r – 1)
AND
r = 0.80
u6 = 1.6 x 0.85
S7 = 1.6 (0.8 7 – 1)
(0.8 – 1)
u6 = 0.524 m
S7 = 6.32 m