Question

determinati ratia progresiei aritmetice 2a10=a5+a6+36

Answer 1

a10=a1+9r
a5=a1+4r
a6=a1+5r înlocuim
2a1+18 r=2a1+9r+36
9r=36
r=4

Answer 2

$\displaystyle 2a_{10}=a_5+a_6+36 \\ 2(a_{10-1}+r)=a_{5-1}+r+a_{6-1}+r+36 \\ 2(a_9+r)=a_4+r+a_5+r+36 \\ 2(a_1+9r)=a_1+4r+a_1+5r+36 \\ 2a_1+18r=a_1+4r+a_1+5r+36 \\ 2a_1-a_1-a_1+18r-4r-5r=36 \\ 18r-4r-5r=36 \\ 9r=36 \Rightarrow r= \frac{36}{9} \Rightarrow \boxed{r=4}$