Question

Sa se determine primul termen si ratia unei progresi aritmetice , stiind ca : a4 + a6 = 6√2 si a5 - a7 = -2√2 . MULTUMESC

Answer 1

a₄=a₁+3r
a₆=a₁+5r
a₄+a₆=a₁+3r+a₁+5r=2a₁+8r = 6√2 
a₅=a₁+4r
a₇=a₁+6r
a₅-a₇=a₁+4r-a₁-6r=-2r=-2√2     ⇒ r=√2

2a₁+8r = 6√2 
2a₁+8√2 = 6√2 
2a₁= 6√2 -8√2=-2√2
a₁=-√2

deci a₁=-√2
         r=√2

Answer 2

$\displaystyle a_4+a_6=6 \sqrt{2} \Rightarrow a_{4-1}+r+a_{6-1}+r=6 \sqrt{2} \Rightarrow \\ \\ \Rightarrow a_3+r+a_5+r=6 \sqrt{2} \Rightarrow a_1+3r+a_1+5r=6 \sqrt{2} \Rightarrow \\ \\ \Rightarrow 2a_1+8r=6 \sqrt{2} \\ \\ a_5-a_7=-2 \sqrt{2} \Rightarrow a_{5-1}+r-(a_{7-1}+r)=-2 \sqrt{2} \Rightarrow \\ \\ \Rightarrow a_4+r-(a_6+r)=-2 \sqrt{2} \Rightarrow a_1+4r-(a_1+6r)=-2 \sqrt{2} \Rightarrow$

$\displaystyle \Rightarrow a_1+4r-a_1-6r=-2 \sqrt{2} \Rightarrow 4r-6r=-2 \sqrt{2} \Rightarrow -2r=-2 \sqrt{2} \Rightarrow \\ \\ \Rightarrow r= \frac{-2 \sqrt{2} }{-2} \Rightarrow \boxed{r= \sqrt{2}} \\ \\ 2a_1+8r=6 \sqrt{2} \Rightarrow 2a_1+8 \sqrt{2} =6 \sqrt{2} \Rightarrow 2a_1=6 \sqrt{2} -8 \sqrt{2} \Rightarrow \\ \\ \Rightarrow 2a_1=-2 \sqrt{2} \Rightarrow a_1=- \frac{2 \sqrt{2} }{2} \Rightarrow \boxed{a_1=- \sqrt{2} }$