sum of integers from 1000 to 2000 not multiples of 7 is the sum of integergs from 1000 to 2000 minus the multiples of seven from1000 to 2000.

find # of integers from 1000 to 2000:     \(2000 = 1000 + \left( {n - 1} \right) \cdot 1\;\;\;\; \Rightarrow \;\;\;\;1000 = n - 1\;\;\;\; \Rightarrow \;\;\;\;n = 1001\)

\({S_{1001}} = \frac{{1001}}{2}\left( {1000 + 2000} \right) = 1\;501\;500\)

find # of multiples of 7 from 1000 to 2000:

\(\frac{{1000}}{7} \approx 142.86\;\;\;\; \Rightarrow \;\;\;\;7 \cdot 143 = 1001\)        \(\frac{{2000}}{7} \approx 285.71\;\;\;\; \Rightarrow \;\;\;\;7 \cdot 285 = 1995\)

\(1995 = 1001 + \left( {n - 1} \right) \cdot 7\;\;\; \Rightarrow \;\;\;994 = 7n - 7\;\;\; \Rightarrow \;\;\;7n = 1001\;\;\; \Rightarrow \;\;\;n = 143\)

\({S_{143}} = \frac{{143}}{2}\left( {1001 + 1995} \right) = 214\;214\)

sum of integers 1000 to 2000 minus multiples of 7 from 1000 to 2000 \(=1501500 - 214214=1287286\)

\(\frac{a}{{\;\frac{1}{4}}} = \frac{9}{a}\;\;\;\; \Rightarrow \;\;\;\;{a^2} = \frac{9}{4}\;\;\;\; \Rightarrow \;\;\;\;a = \frac{3}{2}\)

\(a - \frac{1}{4} = 9 - b - a\;\;\;\; \Rightarrow \;\;\;\;b = \frac{{37}}{4} - 2a\;\;\;\; \Rightarrow \;\;\;\;b = \frac{{37}}{4} - 2\left( {\frac{3}{2}} \right) = \frac{{37}}{4} - \frac{{12}}{4} = \frac{{25}}{4}\)

\({u_1} = 3,\;d = 4\)

\({u_n} = {u_1} + (n - 1)d\;\;\; \Rightarrow \;\;\;399 = 3 + (n - 1)4\;\;\; \Rightarrow \;\;\;n = \frac{{396}}{4} + 1 = 100\)

\({S_n} = \frac{n}{2}({u_1} + {u_n}) = \frac{{100}}{2}(3 + 399) = 20100\)

(a) \({S_{20}} = \frac{{20}}{2}\left[ {2\left( { - 7} \right) + \left( {20 - 1} \right)d} \right]\)

\(620 = 10\left( { - 14 + 19d} \right)\)

\(190d = 760\)

\(d = \frac{{760}}{{190}} = 4\)

(b) \({u_{78}} = - 7 + \left( {78 - 1} \right)4\)

\({u_{78}} = - 7 + 308 = 301\)

\({u_1} = 108,\;{u_n} = 495\),  \(108 = 9 \cdot 12;\;\;\;495 = 9 \cdot 55\)

\({S_n} = \frac{n}{2}\left( {{u_1} + {u_n}} \right)\), need to find \(n\)

\({u_n} = {u_1} + \left( {n - 1} \right)d\;\;\; \Rightarrow \;\;\;495 = 108 + \left( {n - 1} \right)9\;\;\; \Rightarrow \;\;\;9n = 396\;\;\; \Rightarrow \;\;\;n = 44\)

\({S_{44}} = \frac{{44}}{2}\left( {108 + 495} \right) = 13\,266\)

\({u_1} = 2\) and \({S_\infty } = \frac{{{u_1}}}{{1 - r}} = \frac{2}{{1 - r}} = 3\;\;\;\; \Rightarrow \;\;\;\;1 - r = \frac{2}{3}\;\;\;\; \Rightarrow \;\;\;\;r = \frac{1}{3}\)

\({u_n} = {u_1}{r^{n - 1}}\);      \({u_4} = 2{\left( {\frac{1}{3}} \right)^3} = \frac{2}{{27}}\)

\({S_\infty } = \frac{{{u_1}}}{{1 - r}} = \frac{9}{4}\;\;\;\; \Rightarrow \;\;\;\;{u_1} = \frac{9}{4}\left( {1 - r} \right)\)

\({u_1} + {u_1}r + {u_1}{r^2} + {u_1}{r^3} = \frac{{20}}{9}\;\;\; \Rightarrow \;\;\;\frac{9}{4}\left( {1 - r} \right) + \frac{9}{4}\left( {1 - r} \right)r + \frac{9}{4}\left( {1 - r} \right){r^2} + \frac{9}{4}\left( {1 - r} \right){r^3} = \frac{{20}}{9}\)

multiply both sides by \(\frac{4}{9}\): \(1 - r + \left( {1 - r} \right)r + \left( {1 - r} \right){r^2} + \left( {1 - r} \right){r^3} = \frac{{80}}{{81}}\)

\(1 - r + r - {r^2} + {r^2} - {r^3} + {r^3} - {r^4} = \frac{{80}}{{81}}\;\;\;\; \Rightarrow \;\;\;\;1 - {r^4} = \frac{{80}}{{81}}\)

\({r^4} = \frac{1}{{81}}\;\;\;\; \Rightarrow \;\;\;\;r = \pm \,\frac{1}{3}\)   given that \(r\)  is negative; thus \(r = - \frac{1}{3}\)

then \({u_1} = \frac{9}{4}\left( {1 + \frac{1}{3}} \right) = 3\)

(a) probability Alice wins on second kick \( = \left( {0.3} \right)\left( {0.2} \right)\left( {0.7} \right) = 0.042\)

(b) probability Beatriz wins on 3^rd kick\( = \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) = 0.0000864\)

(c) probability that Beatriz wins a game is equal to probability that she wins on 1st kick plus probability that she wins on 2nd kick plus probability that she wins on 3rd kick, etc. 

probability B wins game \( = \left( {0.3} \right)\left( {0.8} \right) + \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) + \left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.2} \right)\left( {0.3} \right)\left( {0.8} \right) + \; \cdots \)

\( = \left( {0.3} \right)\left( {0.8} \right) + {\left( {0.3} \right)^2}\left( {0.2} \right)\left( {0.8} \right) + {\left( {0.3} \right)^3}{\left( {0.2} \right)^2}\left( {0.8} \right) + \; \cdots \)

this is an infinite geometric series with \({u_1} = \left( {0.3} \right)\left( {0.8} \right) = 0.24\), and \(r = \left( {0.3} \right)\left( {0.2} \right) = 0.06\).

The probability that Beatriz wins a game is the sum of this infinite geometric series.

Thus, probability Beatriz wins a game is \({S_\infty } = \frac{{0.24}}{{1 - 0.06}} \approx 0.255\)  [exact value is 12/47]

(d) probability that Alice wins a game is the complement of Beatriz’ probability of winning.

thus, probability that Alice wins a game is \(1 - 0.25531915 \approx 0.745\)                                                                                                                                 [exact value is 35/47]

In a geometric sequence, the ratio of a term and previous term is constant (i.e. common ratio \(r\))

 \(r = \frac{n}{{n - 2}} = \frac{{n + 3}}{n}\;\; \Rightarrow \;\;{n^2} = \left( {n - 2} \right)\left( {n + 3} \right)\)

\({n^2} = {n^2} + n - 6\;\; \Rightarrow \;\;n - 6 = 0\;\; \Rightarrow \;\;n = 6\)

\({u_1} = 6 - 2 = 4\) and \({u_2} = 6\); hence, \(r = \frac{6}{4} = \frac{3}{2}\)

Thus, \({S_5} = \frac{{4\left( {{{\left( {\frac{3}{2}} \right)}^5} - 1} \right)}}{{\frac{3}{2} - 1}} = \frac{{211}}{4}\)

\({u_1} = 100\) and the last term \({u_n} = 300\); find \(n\), the number of terms in the sequence

\({u_n} = 300 = 100 + \left( {n - 1} \right)5\;\; \Rightarrow \;\;n - 1 = \frac{{200}}{5}\;\;\; \Rightarrow \;\;n = 41\)

\({S_{41}} = \frac{{41}}{2}\left( {100 + 300} \right) = 8200\)