گام به گام تمرین صفحه 24 درس 1 ریاضی (1) (مجموعه، الگو و دنباله)

\(\left\{ \begin{array}{l}d = 7\\\\{t_{21}} = 143\end{array} \right.\)

دنباله حسابی نمی باشد.

\(\left\{ \begin{array}{l}d = \sqrt 3 \\\\{t_{21}} = 21\sqrt 3 \end{array} \right.\)

\(\left\{ \begin{array}{l}d = - 3\\\\{t_{21}} = - 50\end{array} \right.\)

\(\left\{ \begin{array}{l}d = \frac{1}{5}\\\\{t_{21}} = \frac{{22}}{5}\end{array} \right.\)

\(\left\{ \begin{array}{l}d = 0\\\\{t_{21}} = 2\end{array} \right.\)

\(\begin{array}{l}{t_n} = {t_1} + d(n - 1)\\\\\left\{ \begin{array}{l}{t_3} = {t_1} + 2d = 20\\\\{t_7} = {t_1} + 6d = 56\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{t_1} = 2\\\\d = 9\end{array} \right. \Rightarrow {t_n} = 2 + 9(n - 1)\\\\ \Rightarrow {t_n} = 9n - 7\end{array}\)

\(\begin{array}{l}{t_n} = {t_1} + d(n - 1)\\\\\left\{ \begin{array}{l}{t_1} + {t_2} + {t_3} = 3{t_1} + 3d = 3\\\\{t_4} + {t_5} + {t_6} = 3{t_1} + 12d = 39\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{t_1} = - 3\\\\d = 4\end{array} \right.\\\\ \Rightarrow {t_n} = - 3 + 4(n - 1) \Rightarrow {t_n} = 4n - 7\end{array}\)

نوع دنباله، دنباله حسابی است، زیرا اختلاف هر دو جمله متوالی یک مقدار ثابت است.

\(\left\{ \begin{array}{l}{t_1} = 1\\\\d = 4\end{array} \right. \Rightarrow {t_n} = 1 + 4(n - 1) \Rightarrow {t_n} = 4n - 3\)

\({t_{}} = 397 \Rightarrow 4m - 3 = 397 \Rightarrow m = 100\)

\(\left. \begin{array}{l}\left\{ \begin{array}{l}{t_1} = 5\\\\{t_3} = 11\end{array} \right. \Rightarrow d = \frac{{{t_3} - {t_1}}}{{3 - 1}} = \frac{{11 - 5}}{2} = 3\\\\ \Rightarrow {t_2} = {t_1} + d = 8\end{array} \right\} \Rightarrow 5\,,\,8\,,\,11\)

\(\left. \begin{array}{l}\left\{ \begin{array}{l}{t_1} = 20\\\\{t_3} = 30\end{array} \right. \Rightarrow d = \frac{{{t_3} - {t_1}}}{{3 - 1}} = \frac{{30 - 20}}{2} = 5\\\\ \Rightarrow {t_2} = {t_1} + d = 25\end{array} \right\} \Rightarrow 20\,,\,25\,,\,30\)

واسطه حسابی بین دو عدد، میانگین آن دو عدد است. به عبارتی دیگر اگر c ، b ، a دنباله حسابی باشند، داریم:

\(b = \frac{{a + c}}{2}\)

\(\begin{array}{l}{t_n} = {t_1} + d(n - 1)\\\\{t_1}\,,\,{t_2}\,,\,{t_3}\,,\,{t_4}\,,\,{t_5}\\\\\left\{ \begin{array}{l}{t_1} + {t_2} + {t_3} + {t_4} + {t_5} = 100 \Rightarrow 5{t_1} + 5d = 100\\\\{t_1} + {t_2} = \frac{{{t_3} + {t_4} + {t_5}}}{3} \Rightarrow {t_1} - 2d = 0\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l}{t_1} = 10\\\\d = 5\end{array} \right. \Rightarrow {t_n} = 10 + 5(n - 1) \Rightarrow {t_n} = 5n + 5\\\\ \Rightarrow \,10\,,\,15\,,\,20\,,\,25\,,\,30\end{array}\)