Question

Exercițiul 4. Răspuns : a1=0, r=7

Answer 1

Explicație pas cu pas:

$\left \{ {{a_{1} + a_{2} + a_{3} = 21} \atop {a_{2} + a_{3} + a_{4} = 42}} \right. \iff \left \{ {{a_{1} + a_{1} + r + a_{1} + 2r = 21} \atop {a_{1} + r + a_{1} + 2r + a_{1} + 3r = 42}} \right.$

$\left \{ {{3a_{1} + 3r = 21} \atop {3a_{1} + 6r = 42}} \right. \iff \left \{ {{a_{1} + r = 7} \atop {a_{1} + 2r = 14}} \right.$

$2r - r = 14 - 7 \implies \bf r = 7$

$a_{1} + 7 = 7 \implies \bf a_{1} = 0$

Answer 2

$\left\{\begin{array}{ccc}a_1+a_2+a_3=21\\a_2+a_3+a_4=42\\\end{array}\right.\Rightarrow \\\\\\ \Rightarrow \left\{\begin{array}{ccc}a_1+a_1+(2-1)\cdot r+a_1+(3-1)\cdot r=21\\a_1+(2-1)\cdot r+a_1+(3-1)\cdot r+a_1+(4-1)\cdot r=42\\\end{array}\right.\Rightarrow$

$\Rightarrow \left\{\begin{array}{ccc}a_1+a_1+r+a_1+2r=21\\a_1+r+a_1+2r+a_1+3r=42\\\end{array}\right.\Rightarrow \left\{\begin{array}{ccc}3a_1+3r=21|\cdot(-1)\\3a_1+6r=42\\\end{array}\right.\Rightarrow$

$\Rightarrow \left\{\begin{array}{ccc}-3a_1-3r=-21\\3a_1+6r=42\\\end{array}\right.\\~~~~~~----------\\~~~~~~~~~~~~/~~~~~~3r=21\Rightarrow r=\dfrac{21}{2} \Rightarrow \boxed{r=7}\\\\r=7\Rightarrow 3a_1+6r=42\Rightarrow 3a_1+6\cdot 7=42\Rightarrow 3a_1+42=42\Rightarrow \\\\\Rightarrow 3a_1=42-42\Rightarrow 3a_1=0\Rightarrow a_1=\dfrac{0}{3} \Rightarrow \boxed{a_1=0}$