Question

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Answer 1

 

$\displaystyle\bf\\53.~\textbf{Intr-o progresie aritmetica}~(a_n)_{n\geq 1}~\textbf{se stie ca ratia este }~q = 2,\\\textbf{iar suma primilor n termeni }~S_n=3069,~unde~n \in N, n\geq 3.\\Determinati~a_8.\\\\a_n=progresie~geometrica\\q=2\\S_n=3069\\\\Rezolvare:\\Folosim formula:\\\\S_n=a_1\cdot\frac{q^n-1}{q-1}\\\\\\a_1\cdot\frac{q^n-1}{q-1}=3069~~~unde~~q=2\\\\\\a_1\cdot\frac{2^n-1}{2-1}=3069\\\\\\a_1\cdot\frac{2^n-1}{1}=3069$

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$\displaystyle\bf\\a_1(2^n-1)=3069\\\\\text{Avem o ecuatie cu 2 necunoscute.}\\\text{Deoarece }n\in N~\text{avem o sansa sa avem un numar finit de solutii.}\\\\\text{Descompunem in factori primi numarul 3069.}\\\\3069=9\times341=9\times11\times31=\boxed{\bf3^2\times11\times31}\\\\\textbf{Calculam numarul de divizori ai lui 3069}\\\\n=(2+1)(1+1)(1+1)=3\times2\times2=12~divizori.$

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$\displaystyle\bf\\D_{3069}=\{1;~3;~9;~11;~31;~33;~93;~99;~279;~341;~1023;~3069\}\\\\a_1(2^n-1)=3069\\\\\implies~(2^n-1)\in D_{3069}\\\\2^n-1=d~~unde~~d\in D_{3069}\\\\\boxed{\bf2^n=d+1}~~unde~n=numarul~de~termeni~(n\geq3)$

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$\displaystyle\bf\\\textbf{Cautam divizori la care daca adaugam 1 devine o putere a lui 2.}\\1+1=2=2^1~~Solutie~respinsa~deoarece~1<3\\3+1=4=2^2~~Solutie~respinsa~deoarece~2<3\\\boxed{\bf31+1=32=2^5\implies n=5}\\\boxed{\bf1023+1=1024=2^{10}\implies n=10}\\\\Solutia~1:\\\\a_1(2^5-1)=3069\\\\a_1=\frac{3069}{2^5-1}=\frac{3069}{32-1}=\frac{3069}{31}\\\\ \boxed{\bf~a_1=99}\\\\a_8=a_1\times q^7=99\times 2^7=99\times128\\\\\boxed{\bf~a_8=12672}$

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$\displaystyle\bf\\Solutia~2:\\\\a_1(2^10-1)=3069\\\\a_1=\frac{3069}{2^5-1}=\frac{3069}{1024-1}=\frac{3069}{1023}\\\\ \boxed{\bf~a_1=3}\\\\a_8=a_1\times q^7=3\times 2^7=3\times128\\\\\boxed{\bf~a_8=384}$