Question
Consider the following sequence of figures.
Figure 1 contains 6 line segments.
(a) Given that Figure n contains 101 line segments, show that n = 20.
(b) Find the total number of line segments in the first 20 figures.
Answer/Explanation
Ans:
(a) We have an arithmetic sequence with u1 = 6, u2 = 11 and u3 = 16.
The common difference is
D = u2 – u1
= 11 – 6
= 5
Hence, if we substitute un = 101, u1 = 6 and d = 5 in un = u1 + (n -1)d and solve the resulting equation for n, we get
101 = 6 + (n – 1) (5)
101 = 5n + 1
100 = 5n
N = 20
(b) Using the sum of n terms formula \(S_{n}=\frac{n}{2}(u_{1}+u_{n})\) with n = 20, we obtain
\(S_{20}=\frac{20}{2}(u_{1}+u_{20})\)
\(S_=\frac{20}{2}(6+101)\)
= 10(107)
=1070
Question
A box contains 8 green and 4 blue marbles. Two marbles are selected at random (without replacement). Find the probability that selected marbles are:
(a) of the same colour;
(b) of different colours.
Answer/Explanation
Ans:
(a) We have
P(of the same colour) = P(both green) + P(both blue)
\( =\frac{8}{12}\times \frac{7}{11}+\frac{4}{12}\times \frac{3}{11}\)
\( =\frac{17}{33}\)
(b) Using the complementary events formula, we get
P(of different colours) = 1 – P(of the same colour)
\( =1-\frac{17}{33}\) [by part (a)]
\( =\frac{16}{33}\)