Question

Fie an progresi aritmetica ax a1=3,a10=12.se cere a) rația b) S57

Answer 1

a. an=a1+(n-1)r
a10=a1+(10-1)r
12=3+9r, 9r=9, r=1
b S57=(a1+a57)n/2
a57=a1+(57-1)r=3+56=59
S57=(3+59)x57/2=62x57/2=31x57=1767

Answer 2

$\displaystyle a_1=3,~a_{10}=12,~r=?,~S_{57}=? \\ \\ a_{10}=12 \Rightarrow a_{10-1}+r=12 \Rightarrow a_9+r=12 \Rightarrow a_1+9r=12 \Rightarrow \\ \\ \Rightarrow 3+9r=12 \Rightarrow 9r=12-3 \Rightarrow 9r=9 \Rightarrow r= \frac{9}{9} \Rightarrow \boxed{r=1} \\ \\ S_{57}= \frac{2 \cdot 3+(57-1) \cdot 1}{2} \cdot 57 \\ \\ S_{57}= \frac{6+56 \cdot 1}{2} \cdot 57 \\ \\ S_{57}= \frac{6+56}{2} \cdot 57 \\ \\ S_{57}= \frac{62}{2} \cdot 57 \\ \\ S_{57}=31 \cdot 57 \\ \\ \boxed{S_{57}=1767}$