Sequence and Series - SS2 Mathematics Past Questions and Answers - page 1
Find the 13th term in the AP \(5,\ 8,\ 11,\ 14,\ \ldots\)
20
31
27
35
\(T_{n} = a + (n - 1)d\)
\(T_{13} = 5 + (13 - 1)3\ \)
\(T_{13} = 5 + (12)3\)
\(T_{13} = 5 + 26\)
\(T_{13} = 31\)
Write in simplest form the \(n\)th term of the sequence \(13,\ 8,\ 3,\ \ldots\)
18 - 5n
\(\frac{18}{5n}\)
\(\frac{- 18}{5n}\)
\(\frac{13 - 5n}{n + 5}\)
\(T_{n} = a + (n - 1)d\)
\(T_{n} = 13 + (n - 1)( - 5)\)
\(T_{n} = 13 - 5n + 5\)
\(T_{n} = 13 + 5 - 5n\)
\(T_{n} = 18 - 5n\)
What is the sum of the first \(15\) terms of the sequence \(- 9,\ - 6,\ - 3,\ \ldots\)
450
- 450
- 180
180
\[S_{n} = \frac{n}{2}\lbrack 2a + (n - 1)d\rbrack\]
\[ {S_{15} = \frac{15}{2}\lbrack 2( - 9) + (15 - 1)(3)\rbrack}\]
\[S_{15} = \frac{15}{2}\lbrack - 18 + (14)(3)\rbrack\]
\(S_{15} = \frac{15}{2}\lbrack - 18 + 42\)]
\[S_{15} = \frac{15}{2}\lbrack 24\rbrack\]
\(S_{15} = 180\)
Find the geometric mean of \(3\) and \(48\)
11
12
13
14
\(Geometric\ mean,\ GM\ = \ \sqrt{al} = \sqrt{3 \times 48} = \ \sqrt{144} = 12\)
Find the first term of a GP if the third term is \(72\) and the sixth term is \(243\).
\(21\frac{1}{3}\)
\(19\frac{2}{3}\)
\(31\frac{1}{3}\)
21
Assume the 3rd term is the geometric mean of the 1st and 6th terms
\[Geometric\ mean,\ GM\ = \ \sqrt{al}\]
\[72 = \sqrt{a \times 243}\]
\[72^{2} = 243a\]
\(a = \frac{72^{2}}{243} = 21.3\)
Using the formula of compounding (a GP) [ \(A = P{(1 + \frac{R}{100})}^{n}\) ] Calculate the population of the world in \(2010\) if the population in \(1990\) was \(5250\) million and the world experiences a growth rate of \(1.6\%\).
\[A = 5250{(1 + \frac{1.6}{100})}^{20}\]
\[A = 5250{(1 + 0.016)}^{20}\]
\[A = 5250{(1.016)}^{20}\]
\[A = 7212\]
Hence, there will be an estimated \(7212\) million people in 2010.
A man’s annual salary is \(\$ 30\) and after \(10\) years of service, he has a cumulative salary of \(\$ 1014\), find his initial salary.
\[T_{n} = a + (n - 1)d\]
\[T_{10} = a + (10 - 1)1014\]
\[1014 = a + (10 - 1)30\]
\[1014 = a + (9)30\]
\[1014 = a + 270\]
\[1014 - 270 = a\]
\[a = \$ 744\]
If \(2^{n - 1}\) is the \(n\)th term of a GP. Write out the first three terms of the sequence.
\[T_{n} = 2^{n - 1}\]
\[T_{1} = 2^{1 - 1} = 2^{0} = 1\]
\[T_{2} = 2^{2 - 1} = 2^{1} = 2\]
\[T_{3} = 2^{3 - 1} = 2^{2} = 4\]
The first three term of this GP are \(1,\ 2\ and\ 4\)
Check for convergence and the sum to infinity of the series \(1 - 0.1 + 0.01 - 0.001 + \ldots\)
\(Given\ 1 - 0.1 + 0.01 - 0.001 + \ldots\), \(r = \frac{0.01}{- 0.1} = - \frac{1}{10}\), \(|r| = \frac{1}{10} < 1\)
This series converges
\[S_{\infty} = \frac{1}{1 - ( - 0.1)} = \frac{1}{1 + 0.1} = \frac{1}{1.1} = 0.909090\ldots\ \approx 0.91\]