دنباله ي هندسي

چون$a = 2$ و$q = \frac{1}{2}$ پس${a_n} = 2{\left( {\frac{1}{2}} \right)^{n - 1}}$كه طبق فرض بايد$2{\left( {\frac{1}{2}} \right)^{n - 1}} > {10^{ - 3}}$باشد بنابراين:

$\begin{array}{l}2 \times {\left( {\frac{1}{2}} \right)^n}{\left( {\frac{1}{2}} \right)^{ - 1}} > \frac{1}{{1000}} \Rightarrow {\left( {\frac{1}{2}} \right)^n} > \frac{1}{{4000}} \Rightarrow {2^n} < 4000 \Rightarrow n < 12\\\end{array}$

درنتيجه يازده جمله بزرگتر از$\frac{1}{{1000}}$دارد.

$\begin{array}{l}\frac{{{a_5}}}{{{a_3}}} = {q^3} \Rightarrow \frac{{96}}{{12}} = {q^3} \Rightarrow q = 2\\\\\frac{{{a_7}}}{{{a_5}}} = {q^3} \Rightarrow {a_7} = 96 \times 4 \Rightarrow {a_7} = 384\end{array}$

${a_1}{a_2}....{a_9} = {({a_5})^9} = {1^9} = 1$

$\begin{array}{l}\frac{{{s_6}}}{{{s_3}}} = \frac{{63}}{7}\\\\ \Rightarrow \frac{{\frac{{a(1 - {q^6})}}{{1 - q}}}}{{\frac{{a(1 - {q^3})}}{{1 - q}}}} = 9 = \frac{{1 - {q^6}}}{{1 - {q^3}}} = 9\\\\ \Rightarrow 1 + {q^3} = 9 \Rightarrow {q^3} = 8 \Rightarrow q = 2\\\\ \Rightarrow \frac{{{a_6}}}{{{a_3}}} = {q^4} \Rightarrow \frac{{{a_6}}}{{{a_2}}} = 16\end{array}$

چون $a = 1$ و $q = \frac{1}{3}$و ميخواهيم${s_n} < 1/497$باشد پس داريم:

$\begin{array}{l}\frac{{1\left( {1 - {{(\frac{1}{3})}^n}} \right)}}{{1 - \frac{1}{3}}} < 1/497 \Rightarrow \frac{3}{2}\left( {1 - {{\left( {\frac{1}{3}} \right)}^n}} \right) < \frac{{1497}}{{1000}} \Rightarrow 1 - {\left( {\frac{1}{3}} \right)^n} < \frac{{499}}{{500}} \Rightarrow {\left( {\frac{1}{3}} \right)^n} > \frac{1}{{500}} \Rightarrow \\\\{3^n} < 500 \Rightarrow n < 6\end{array}$

$a = 1$

$\left| q \right| < 1 \Rightarrow s = \frac{1}{{1 - ( - \frac{1}{4})}} = \frac{1}{{\frac{5}{4}}} = \frac{4}{5}$

 اگر جمله اول a و قدر نسبت q باشد داريم:

$a = \frac{{aq}}{{1 - q}} \Rightarrow 1 = \frac{1}{2} \Rightarrow \frac{{{a_5}}}{{{a_1}}} = {q^4} = \frac{1}{{16}}$