Question

Sa se scrie termenii : a2 , a5, a12, a26 ai progresiei aritmetice (an) daca ​

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle a_n=a_1+(n-1)r \\\\a)~a_1=-3,~r=2\\\\ a_2=-3+(2-1) \cdot 2=-3+1 \cdot 2=-3+2=-1\\\\ a_5=-3+(5-1)\cdot 2=-3+4 \cdot 2=-3+8=5\\\\ a_{12}=-3+(12-1) \cdot 2=-3+11 \cdot 2=-3+22=19\\\\ a_{26}=-3+(26-1) \cdot 2=-3+25 \cdot 2=-3+50=47$

$b)~a_{15}=49,~r=3 \\\\ a_{15}=a_1+(15-1) \cdot r=a_1+14r\\\\ a_{15}=49 \Rightarrow a_1+14r=49 \Rightarrow a_1+14 \cdot 3=49 \Rightarrow a_1+42=49 \Rightarrow \\\\ \Rightarrow a_1=49-42\Rightarrow a_1=7\\\\a_2=7+(2-1) \cdot 3=7+1 \cdot 3=7+3=10\\\\ a_5=7+(5-1) \cdot 3=7+4 \cdot 3=7+12=19\\\\ a_{12}=7+(12-1) \cdot 3=7+11 \cdot 3=7+33=40\\\\ a_{26}=7+(26-1) \cdot 3=7+25 \cdot 3=7+75=82$

$c)~a_{30}=97,~a_{31}=102 \\\\ \left\{\begin{array}{ccc}a_{30}=97\\a_{31}=102\end{array}\right \Rightarrow \left\{\begin{array}{ccc}a_1+(30-1) \cdot r=97\\a_1+(31-1)\cdot r=102 \end{array}\right \Rightarrow \left\{\begin{array}{ccc}a_1+29r=97\big|\cdot(-1)\\a_1+30r=102\end{array}\right \Rightarrow \\\\ \Rightarrow \left\{\begin{array}{ccc}-a_1-29r=-97\\a_1+30r=102\end{array}\right \\~~~~----------\\~~~~~~~~~~~/~~~~~~~r=5$

$\displaystyle a_1+29r=97 \Rightarrow a_1+29 \cdot 5=97 \Rightarrow a_1+145=97 \Rightarrow a_1=97-145\Rightarrow \\\\ \Rightarrow a_1=-48\\\\ a_2=-48+(2-1) \cdot 5=-48+1 \cdot 5=-48+5=-43\\\\ a_5=-48+(5-1)\cdot5=-48+4 \cdot 5=-48+20=-28\\\\ a_{12}=-48+(12-1) \cdot 5=-48+11 \cdot 5=-48+55=7\\\\ a_{26}=-48+(26-1)\cdot 5=-48+25 \cdot 5=-48+125=77$

$\displaystyle d)~a_5=16,~a_9=64\\\\ \left\{\begin{array}{ccc}a_5=16\\a_9=64\end{array}\right \Rightarrow \left\{\begin{array}{ccc}a_1+(5-1)\cdot r=16\\a_1+(9-1)\cdot r=64\end{array}\right\Rightarrow \left\{\begin{array}{ccc}a_1+4r=16\big|\cdot(-1)\\a_1+8r=64\end{array}\right \Rightarrow \\\\ \Rightarrow \left\{\begin{array}{ccc}-a_1-4r=-16\\a_1+8r=64\end{array}\right\\~~~~----------\\~~~~~~~~~~~/~~~~~4r=48 \Rightarrow r=\frac{48}{4} \Rightarrow r=12$

$a_1+4r=16 \Rightarrow a_1+4 \cdot 12=16 \Rightarrow a_1+48=16 \Rightarrow a_1=16-48 \Rightarrow a_1=-32\\\\ a_2=-32+(2-1) \cdot 12=-32+1 \cdot 12=-32+12=-20\\\\ a_5=-32+(5-1)\cdot 12=-32+4 \cdot 12=-32+48=16\\\\ a_{12}=-32+(12-1) \cdot 12=-32+11 \cdot 12=-32+132=100\\\\ a_{26}=-32+(26-1) \cdot 12=-32+25\cdot 12=-32+300=268$