IB Mathematics SL 1.3 Geometric sequences and series AI HL Paper 2- Exam Style Questions- New Syllabus

Question

Scott purchases food for his dog in large bags and feeds the dog the same amount of dog food each day. The amount of dog food left in the bag at the end of each day can be modelled by an arithmetic sequence.

On a particular day, Scott opened a new bag of dog food and fed his dog. By the end of the third day there were 115.5 cups of dog food remaining in the bag and at the end of the eighth day there were 108 cups of dog food remaining in the bag.

(a) (i) Find the number of cups of dog food fed to the dog per day:
(ii) Find the number of cups of dog food remaining in the bag at the end of the first day:

(b) Calculate the number of days that Scott can feed his dog with one bag of food:

(c) In 2021, Scott spent $625 on dog food. Scott expects that the amount he spends on dog food will increase at an annual rate of 6.4%. Determine the amount that Scott expects to spend on dog food in 2025. Round your answer to the nearest dollar:

(d) (i) Calculate the value of $\sum_{n=1}^{10} (625 \times 1.064^{(n-1)})$:
(ii) Describe what the value in part (d)(i) represents in this context:

(e) Comment on the appropriateness of modelling this scenario with a geometric sequence:

▶️ Answer/Explanation

Markscheme

(a)(i)
    Either:
    115.5 = $ u_1 + (3 – 1) \times d $ (115.5 = $ u_1 + 2d $)
    108 = $ u_1 + (8 – 1) \times d $ (108 = $ u_1 + 7d $)
    Subtract: $ 115.5 – 108 = (u_1 + 2d) – (u_1 + 7d) $
    $ 7.5 = -5d $
    $ d = \frac{7.5}{-5} $
    $ d = -1.5 $
    Dog eats $ |-1.5| = 1.5 $ cups/day
    Or:
    $ d = \frac{115.5 – 108}{8 – 3} $
    $ d = \frac{7.5}{5} $
    $ d = 1.5 $ (cups/day)
Result:
1.5 cups/day

(a)(ii)
    Using $ 115.5 = u_1 + 2 \times (-1.5) $
    $ 115.5 = u_1 + (-3) $
    $ u_1 = 115.5 + 3 $
    $ u_1 = 118.5 $ (cups)
Result:
118.5 cups

(b)
    Set bag empty when $ u_n = 0 $
$ 0 = 118.5 + (n – 1) \times (-1.5)$
   $ 0 = 118.5 – 1.5 \times (n – 1)$
$   1.5 \times (n – 1) = 118.5$
    $ n – 1 = \frac{118.5}{1.5} $
    $ n – 1 = 79 $
    $ n = 80 $ days
Result:
80 days

(c)
    Year 2025 is 4 years after 2021
    $ t_5 = 625 \times 1.064^{(5-1)} $
    $ t_5 = 625 \times 1.064^4 $
    $ 1.064^2 \approx 1.131936 $
    $ 1.064^4 \approx 1.131936 \times 1.131936 \approx 1.281 $
    $ 625 \times 1.281 \approx 800.625 $
    Rounded to nearest dollar: $801
Result:
$801

(d)(i)
    Use sum formula for geometric series
    $ S_n = t_1 \frac{1 – r^n}{1 – r} $
    $ S_{10} = 625 \frac{1 – 1.064^{10}}{1 – 1.064} $
    $ 1.064^{10} \approx 1.851 $
    $ 1 – 1.851 = -0.851 $
    $ 1 – 1.064 = -0.064 $
    $ \frac{-0.851}{-0.064} \approx 13.296875 $
    $ 625 \times 13.296875 \approx 8394 $
    Rounded to nearest dollar: $8394
Result:
$8394

(d)(ii)
    Sum from $ n=1 $ (2021) to $ n=10 $ (2030)
    Either:
    the total cost (of dog food) for 10 years beginning in 2021 OR 10 years before 2031
    Or:
    the total cost (of dog food) from 2021 to 2030 (inclusive) OR from 2021 to (the start of) 2031
Result:
the total cost (of dog food) from 2021 to 2030 (inclusive)

(e)
    Geometric sequence assumes constant 6.4% increase
    Either:
    According to the model, the cost of dog food per year will eventually be too high to keep a dog.
    Or:
    The model does not necessarily consider changes in inflation rate.
    Or:
    The model is appropriate as long as inflation increases at a similar rate.
    Or:
    The model does not account for changes in the amount of food the dog eats as it ages/becomes ill/stops growing.
    Or:
    The model is appropriate since dog food bags can only be bought in discrete quantities.
Result:
The model does not account for changes in the amount of food the dog eats as it ages/becomes ill/stops growing.

IB Mathematics SL 1.3 Geometric sequences and series AI HL Paper 2- Exam Style Questions- New Syllabus

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