IBDP Maths AI: Topic : SL 1.2: Arithmetic sequences and series: IB style Questions SL Paper 1

Question 8.[Maximum mark: 8]

Charlie and Daniella each began a fitness programme. On day one, they both ran 500 m. On each subsequent day, Charlie ran 100 m more than the previous day whereas Daniella increased her distance by 2 % of the distance ran on the previous day.

(a) Calculate how far

(i) Charlie ran on day 20 of his fitness programme.

(ii) Daniella ran on day 20 of her fitness programme. [5]

(iii) On day n of the fitness programmes Daniella runs more than Charlie for the first time.

Find the value of n . [3]

Answer/Explanation

(a) (i) u20 using an arithmetic sequence \( u1 = 500\) and \(d = 100\)

   \( u20=  500+ 1900\)  OR 500,600,700,… (Charlie ran) 2400 m

(ii) \(r = 1.02\) 

u20 using a geometric sequence \(u1 = 500\)

\(T_{20}=500 × r19 \)OR

500, 510, 520.2,… (Daniella ran) 728 m (728.405…)

(b) \( 500 × 1.02 n-1 > 500 + (n-1) × 100\) on solving inequality

\(n >184.215… n =185\)

Question

The fifth term of an arithmetic sequence is 20 and the twelfth term is 41.

(i) Find the common difference.

(ii) Find the first term of the sequence.[3]

a.

Calculate the eighty-fourth term.[1]

b.

Calculate the sum of the first 200 terms.[2]

c.
Answer/Explanation

Markscheme

(i) \(u_5 = u_1 + 4d = 20\)

\(u_{12} = u_1 + 11d = 41\)     (M1)

(M1) for both equations correct (or (M1) for \(20 + 7d = 41\))

\(7d = 21\)

\(d = 3\)     (A1)     (C2)

(ii) \(u_1 + 12 = 20\)

\(u_1 = 8\)     (A1)(ft)     (C1) [3 marks]

a.

\(u_{84} = 8 + (84 – 1)3\)

\(= 257\)     (A1)(ft)     (C1)[1 mark]

b.

\(S_{200} = 100(16 + 199 \times 3)\)     (M1)

\( = 61300\)     (A1)(ft)     (C2)[2 marks]

c.

Question

The first term of an arithmetic sequence is \(0\) and the common difference is \(12\).

Find the value of the \({96^{{\text{th}}}}\) term of the sequence.[2]

a.

The first term of a geometric sequence is \(6\). The \({6^{{\text{th}}}}\) term of the geometric sequence is equal to the \({17^{{\text{th}}}}\) term of the arithmetic sequence given above.

Write down an equation using this information.[2]

b.

The first term of a geometric sequence is \(6\). The \\({6^{{\text{th}}}}\) term of the geometric sequence is equal to the \({17^{{\text{th}}}}\) term of the arithmetic sequence given above.

Calculate the common ratio of the geometric sequence.[2]

c.
Answer/Explanation

Markscheme

\({u_{96}} = {u_1} + 95d\)     (M1)

\( = 0 + 95 \times 12\)

\( = 1140\)     (A1)     (C2)[2 marks]

a.

\(6{r^5} = 16d\)     (A1)

\(6{r^5} = 16 \times 12\) (\(192\))     (A1)     (C2)

Note: (A1) only, if both terms seen without an equation.[2 marks]

b.

\({r^5} = 32\)     (A1)(ft)

Note: (ft) from their (b).

\(r = 2\)     (A1)(ft)     (C2)[2 marks]

c.

Question

Given the arithmetic sequence: \({u_1} = 124{\text{, }}{u_2} = 117{\text{, }}{u_3} = 110{\text{, }}{u_4} = 103, \ldots \)

Write down the common difference of the sequence.[1]

a.

Calculate the sum of the first \(50\) terms of the sequence.[2]

b.

\({u_k}\) is the first term in the sequence which is negative.

Find the value of \(k\).[3]

c.
Answer/Explanation

Markscheme

\(d = – 7\)     (A1)     (C1)[1 mark]

a.

\({S_{50}} = \frac{{50}}{2}(2(124) + 49( – 7))\)     (M1)

Note: (M1) for correct substitution.

\( = – 2375\)     (A1)(ft)     (C2)[2 marks]

b.

\(124 – 7(k – 1) < 0\)     (M1)

\(k > 18.7\) or \(18.7\) seen     (A1)(ft)

\(k = 19\)     (A1)(ft)     (C3)

Note: (M1) for correct inequality or equation seen or for list of values seen or for use of trial and error.[3 marks]

c.

Question

A tree begins losing its leaves in October. The number of leaves that the tree loses each day increases by the same number on each successive day.

Calculate the number of leaves that the tree loses on the 21st October.[3]

a.

Find the total number of leaves that the tree loses in the 31 days of the month of October.[3]

b.
Answer/Explanation

Markscheme

\(u_{21} = 24 + (21 – 1)(16)\)     (M1)(A1)

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

\(u_{21} = 344\)     (A1)     (C3)[3 marks]

a.

\({S_{31}} = \frac{{31}}{2}\left[ {2(24) + (31 – 1)(16)} \right]\)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions. (ft) from their value for d.

\(S_{31} = 8184\)     (A1)(ft)     (C3)[3 marks]

b.
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