Question7 [Maximum mark: 6]
A meteorologist models the height of a hot air balloon launched from the ground. The model assumes the balloon travels vertically upwards and travels 450 metres in the first minute.
Due to the decrease in temperature as the balloon rises, the balloon will continually slow down. The model suggests that each minute the balloon will travel only 82 % of the distance travelled in the previous minute.
a. Find how high the balloon will travel in the first 10 minutes after it is launched. [3]
The balloon is required to reach a height of at least 2520 metres.
b. Determine whether it will reach this height. [2]
c. Suggest a limitation of the given model. [1]
▶️Answer/Explanation
(a) recognition of geometric sequence r = 0.82
\(S_{10}= \frac{450(1-0.82^{10})}{1-0.82}\)
\(= 2160 m (2156.37..)\)
(b) \(S=\frac{450}{1-0.82}\)
\(= 2500<2520\) so the balloon will not reach the required height.
(c) horizontal motion not taken into account, rate of cooling will not likely be linear, balloon is considered a point mass / size of balloon not considered, effects of wind/weather unlikely to be consistent, a discrete model has been used, whereas a continuous one may offer greater accuracy
Question 3.[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) attempt to find u20 using an arithmetic sequence e.g. u1 = 500 and d = 100 OR u20= 500+ 1900 OR 500,600,700,… (Charlie ran) 2400 m (ii) (r = ) 1.02 attempt to find u20 using a geometric sequence e.g. identifying u1 = 500 and a value for r OR 500 × r19 OR 500, 510, 520.2,… (Daniella ran) 728 m (728.405…) (b) 500 × 1.02 n-1 > 500 + (n-1) × 100 attempt to solve inequality n >184.215… n =185
Question
The fifth term of an arithmetic sequence is 20 and the twelfth term is 41.
a.(i) Find the common difference.
(ii) Find the first term of the sequence.[3]
b.Calculate the eighty-fourth term.[1]
c.Calculate the sum of the first 200 terms.[2]
▶️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]
\(u_{84} = 8 + (84 – 1)3\)
\(= 257\) (A1)(ft) (C1)[1 mark]
\(S_{200} = 100(16 + 199 \times 3)\) (M1)
\( = 61300\) (A1)(ft) (C2)[2 marks]