Sequence and Series - SS2 Mathematics Past Questions and Answers - page 1

\(T_{n} = a + (n - 1)d\)

\(T_{13} = 5 + (13 - 1)3\ \)

\(T_{13} = 5 + (12)3\)

\(T_{13} = 5 + 26\)

\(T_{13} = 31\)

\(T_{n} = a + (n - 1)d\)

\(T_{n} = 13 + (n - 1)( - 5)\)

\(T_{n} = 13 - 5n + 5\)

\(T_{n} = 13 + 5 - 5n\)

\(T_{n} = 18 - 5n\)

\[S_{n} = \frac{n}{2}\lbrack 2a + (n - 1)d\rbrack\]

\[ {S_{15} = \frac{15}{2}\lbrack 2( - 9) + (15 - 1)(3)\rbrack}\]

\[S_{15} = \frac{15}{2}\lbrack - 18 + (14)(3)\rbrack\]

\(S_{15} = \frac{15}{2}\lbrack - 18 + 42\)]

\[S_{15} = \frac{15}{2}\lbrack 24\rbrack\]

\(S_{15} = 180\)

\(Geometric\ mean,\ GM\ = \ \sqrt{al} = \sqrt{3 \times 48} = \ \sqrt{144} = 12\) 

Assume the 3rd term is the geometric mean of the 1st and 6th terms

\[Geometric\ mean,\ GM\ = \ \sqrt{al}\]

\[72 = \sqrt{a \times 243}\]

\[72^{2} = 243a\]

\(a = \frac{72^{2}}{243} = 21.3\)

\[A = 5250{(1 + \frac{1.6}{100})}^{20}\]

\[A = 5250{(1 + 0.016)}^{20}\]

\[A = 5250{(1.016)}^{20}\]

\[A = 7212\]

Hence, there will be an estimated \(7212\) million people in 2010.

\[T_{n} = a + (n - 1)d\]

\[T_{10} = a + (10 - 1)1014\]

\[1014 = a + (10 - 1)30\]

\[1014 = a + (9)30\]

\[1014 = a + 270\]

\[1014 - 270 = a\]

\[a = \$ 744\]

\[T_{n} = 2^{n - 1}\]

\[T_{1} = 2^{1 - 1} = 2^{0} = 1\]

\[T_{2} = 2^{2 - 1} = 2^{1} = 2\]

\[T_{3} = 2^{3 - 1} = 2^{2} = 4\]

The first three term of this GP are \(1,\ 2\ and\ 4\)

\(Given\ 1 - 0.1 + 0.01 - 0.001 + \ldots\), \(r = \frac{0.01}{- 0.1} = - \frac{1}{10}\), \(|r| = \frac{1}{10} < 1\)

This series converges

\[S_{\infty} = \frac{1}{1 - ( - 0.1)} = \frac{1}{1 + 0.1} = \frac{1}{1.1} = 0.909090\ldots\ \approx 0.91\]