Chapter 19

1.
1. $a_{1} = 8, d = - 1/2;$ arith. seq. 8, 15/2, 7, 13/2, 6, …
2. $a_{1} = 8, r = - 1/2;$ geom. seq. 8, $- 4, 2, - 1, 1/2, \dots$
2. 6, $- 2, \frac{2}{3}, \dots;$

geometric sequence

$\begin{array}{rcl} a_{1} & = & 6 r = \frac{- 2}{6} = - \frac{1}{3} \\ S_{7} & = & \frac{6 [1 - {(- \frac{1}{3})}^{7}]}{1 - (- \frac{1}{3})} \\ = & \frac{6 (1 + \frac{1}{3^{7}})}{\frac{4}{3}} = \frac{1094}{243} \end{array}$
3. $\begin{array}{l} d = (x + 6) - 4 & and & d = (3 x + 2) - (x + 6) \\ d = x + 2 & and & d = 2 x - 4 \end{array}$

$\begin{array}{rcl} So x + 2 & = & 2 x - 4 \\ x & = & 6 \end{array}$

$\begin{array}{rcl} Thus, d & = & 8. \\ a_{10} & = & 4 + 9 (8) = 76 \\ S_{10} & = & \frac{10}{2} (4 + 76) = 400 \end{array}$
4. $\begin{array}{l} 0.454545 \dots = 0.45 + 0.0045 + 0.000045 + \dots \\ a = 0.45 r = 0.01 \\ S = \frac{0.45}{1 - 0.01} = \frac{0.45}{0.99} = \frac{5}{11} \end{array}$
5. $\begin{array}{rcl} \sqrt{1 - 4 x} & = & {(1 - 4 x)}^{1/2} \\ = & 1 + (\frac{1}{2}) (- 4 x) + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2} {(- 4 x)}^{2} + \dots \\ = & 1 - 2 x - 2 x^{2} + \dots \end{array}$
6. $\begin{array}{l} {(2 x - y)}^{5} = {(2 x)}^{5} + 5 {(2 x)}^{4} (- y) + \frac{5 (4)}{2} {(2 x)}^{3} {(- y)}^{2} \\ + \frac{5 (4) (3)}{2 (3)} {(2 x)}^{2} {(- y)}^{3} + \frac{5 (4) (3) (2)}{2 (3) (4)} (2 x) {(- y)}^{4} + {(- y)}^{5} \\ = 32 x^{5} - 80 x^{4} y + 80 x^{3} y^{2} \end{array}$

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Basic Technical Mathematics, 11th Edition by Allyn J. Washington, Richard Evans