Question

Exercițiul 5. Răspuns final :80.

Answer 1

$\it a_6=a_3+3r \Rightarrow 9=3+3r \Big|_{:3} \Rightarrow 3=1+r\Big|_{-1} \Rightarrow r=2\\ \\ \\ a_1=a_3-2r=3-4=-1\\ \\ \\ a_{10}=a_6+4r=9+8=17}\\ \\ \\ S_{10}=\dfrac{(a_1+a_{10})\cdot10}{2}=\dfrac{(-1+17)\cdot10}{2}=16\cdot5=80$

Answer 2

Răspuns:

80

Explicație pas cu pas:

$\begin{cases}a_{3} = 3 \\a_{6} = 9 \end{cases} \iff \begin{cases} a_{1} + 2 \cdot r = 3\\a_{1} + 5 \cdot r = 9 \end{cases}$

$3r = 6 \implies \bf r = 2$

$a_{1} = 3 - 2 \cdot 2 = 3 - 4 \implies \bf a_{1} = - 1$

$S_{n} = \frac{[2a_{1} + (n - 1)\cdot r]\cdot n}{2}$

$S_{10} = \frac{[2 \cdot ( - 1) + (10 - 1)\cdot 2]\cdot 10}{2} = \frac{( - 2 + 18)\cdot 10}{2} = 16\cdot 5 = \bf 80$