جواب کاردرکلاس صفحه 70 درس 2 ریاضی و آمار دوازدهم انسانی (الگوهای خطی)

 الف

\(\begin{array}{l}\left\{ \begin{array}{l}{a_1} = - 1\\{a_2} = 4\end{array} \right.\;\;\; \Rightarrow \;\,{a_2} - {a_1} = d = 5\\\\{a_n} = 199 \Rightarrow {a_n} = {a_1} + d(n - 1)\\\\ \Rightarrow 199 = - 1 + 5(n - 1) \Rightarrow n - 1 = 40\\\\ \Rightarrow n = 41\\\\{S_{41}} = \frac{{41}}{2}( - 1 + 199) = 4059\end{array}\)

ب 

\(\begin{array}{l} - 2\;,\;3\;,\;8\;,\;13\;,\;18\;,\;23\;,\;28\;,\;33\;,\;38\;,\;43\\\\ \Rightarrow \;\;\left\{ \begin{array}{l}{a_1} = - 2\\{a_{10}} = 43\end{array} \right.\quad \\\\ \Rightarrow {S_{10}} = \frac{{10}}{2}(43 + ( - 2)) = 205\end{array}\)

\(\begin{array}{l}{a_1} = 1\;\;\;\;{a_n} = {a_1} + \left( {n - 1} \right)d = 1 + 2\left( {n - 1} \right) = 2n - 1\\\\{a_{30}} = 2 \times 30 - 1 = 59\\\\{S_{30}} = \frac{{30}}{2}\left( {{a_1} + {a_{30}}} \right) = 15\left( {1 + 59} \right) = 15 \times 60 = 900\end{array}\)

\(\begin{array}{l}15\:,\:18\:,\:21\:,\: \cdots \\\\\left\{ {\begin{array}{*{20}{l}}{{S_n} = 870}\\{d = 3}\\{{a_1} = 15}\\{n = ?}\end{array}} \right.\\\\ \Rightarrow {S_n} = \frac{n}{2}(2{a_1} + (n - 1))\\\\ \Rightarrow 870 = \frac{n}{2}(30 + 3n - 3)\\\\\: \Rightarrow 1740 = n(27 + 3n)\\\\ \Rightarrow 3{n^2} + 27n - 1740 = 0\\\\ \Rightarrow {n^2} + 9n - 580 = 0\\\\ \Rightarrow (n - 20)(n + 29) = 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{n = 20}\\{n = - 29\; \otimes }\end{array}} \right.\end{array}\)

چون n باید عدد طبیعی باشد.

با جای گذاری d=3 در رابطه بالا n=20 به دست می آید.