IB Maths AA SL 1.3 Geometric sequences and series Question Bank

Practise IB Maths AA SL Geometric sequences and series with SL questions that develop route-specific sequence or financial reasoning and exam accuracy.

[Maximum number: 17]

Give your answers to parts (a)(ii), (c)(i) and (d) correct to two decimal places.
Daniela and Sorin have each recently received some money. Daniela won a cash prize and Sorin received an inheritance.
Daniela had two options to choose from to receive her winnings. In both options she receives a payment on the first day of each month for three years.
Option A Each payment is $ 5500.
Option B The first payment is $ 2000. In each month which follows, the payment is 6 % more than the previous month.

(a)

Find the total amount Daniela would receive if she chooses

[ 5 ]

(i)

Option B.

Sorin received an inheritance of $ 120000. Sorin invested his inheritance in an account that pays a nominal annual interest rate of 4 % per annum, compounded monthly. The interest is added on the last day of each month.

[ 5 ]

recognizing sum of a geometric sequence is required
$\frac{2000\left(1-1.06^{36}\right)}{1-1.06}$ $=238241.7333 \ldots$
=($) 238241.73

(b)

Find the total value of Daniela's winnings and Sorin's investment on the last day of the sixth month.

[ 6 ]

Sorin's total $=120000\left(1+\frac{4}{100 \times 12}\right)^{6}(=122420.09)$

Daniela's total $=\frac{2000\left(1-1.06^{6}\right)}{1-1.06}(=13950.64)$
total value =($) 136370.73

(c)

Find the minimum number of complete months before the total value of Daniela's winnings and Sorin's investment is at least $\$ 250000$ .

At the end of the three years, Daniela invested $40000 for a further six years in a second account that pays a nominal interest rate of r % per annum compounded quarterly.

[ 6 ]

EITHER (finding number of months, m )
$120000\left(1+\frac{4}{100 \times 12}\right)^{m}+\frac{2000\left(1-1.06^{m}\right)}{1-1.06}(\geq 250000)$ $m \geq 26.0905 \ldots$ OR $(m=26 \Rightarrow) 249157 \ldots$ AND $(m=27 \Rightarrow) 258692 \ldots$

Note: Condone use of an equation or strict inequality.
OR (finding number of years, n )
$120000\left(1+\frac{4}{100 \times 12}\right)^{12 \times n}+\frac{2000\left(1-1.06^{12 \times n}\right)}{1-1.06}(\geq 250000)$ $n \geq 2.17421 \ldots$ (years)

Note: Condone use of an equation or strict inequality.

THEN
m=27 (months)

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SL 1.3—Geometric sequences and series question 1

Question (a)

Question (i)

Question (b)

Question (c)