Răspuns:
9
Explicație pas cu pas:
[tex]S_n = a_1 + a_2 + a_3 + \cdots + a_n\\ \\ \text{Suma urmatorilor n termeni} = a_{n+1} + a_{n+2}+ a_{n+3} + \cdots + a_{2n}\\ \\ \text{Obs: }a_{n+1} + a_{n+2}+ a_{n+3} + \cdots + a_{2n} = S_{2n} - S_n\\ \\ S_n = \frac{1}{3}\cdot (S_{2n} - S_n)\\ \\ S_n = \frac{1}{3}S_{2n} - \frac{1}{3}S_n\\ \\ \frac{4}{3}S_n = \frac{1}{3}S_{2n}\\ \\ 4S_n = S_{2n}\\ \\ S_n = \frac{n(a_1 + a_n)}{2}\\ \\ S_{2n} = \frac{2n(a_1+a_{2n})}{2} = n(a_1+a_{2n})\\ \\ 4\cdot \frac{n(a_1+a_n)}{2} = n(a_1 + a_{2n})\\ \\ 2n(a_1+a_n) = n(a_1 + a_1 + (2n-1)\cdot r)\\ \\ 2\cdot (a_1+a_n) = (a_1 + a_1 + 2nr - r)\\ \\ 2\cdot (a_1+a_n) = (a_1 + a_1 + nr - r + nr)\\ \\ 2\cdot (a_1+a_n) = (a_1 + a_1 + (n - 1)r + nr)\\ \\ 2\cdot (a_1+a_n) = (a_1 + a_n + nr)\\ \\ 2\cdot (a_1+a_n) = (a_1 + a_n) + nr\\ \\ \boxed{(a_1+a_n) = nr}\\ \\ a_1 + a_1 + nr - r = nr \implies 2a_1 - r = 0 \implies \boxed{r = 2a_1} \\ \\ S_{3n} = \frac{3n(a_1+a_{3n})}{2}[/tex]
[tex]\frac{S_{3n}}{S_n} = \dfrac{\frac{3n(a_1+a_{3n})}{2}}{\frac{n(a_1+a_n)}{2}} = \frac{3(a_1 + a_{3n})}{a_1 + a_n}\\ \\ \frac{3(a_1 + a_1 + 3nr - r)}{a_1 + a_n}\\ \\ \frac{3(2a_1 + 3\cdot nr - r)}{a_1 + a_n}\\ \\ \frac{3(2a_1 + 3\cdot (a_1+a_n) - r)}{a_1 + a_n} \\ \\ \frac{3(2a_1 + 3(a_1 + a_n) - 2a_1)}{a_1 + a_n} = \frac{3\cdot 3(a_1+a_n)}{a_1+a_n} = \boxed{9}[/tex]
