Question

vă rog frumos, e urgent, dau coroana!
ex 23.​

Answer 1

Răspuns:

Explicație pas cu pas:

$a_{2}=\sqrt{a_{1}a_{4}},\\{ a_{2}}^{2} ={a_{1}a_{4}} \\{ (a_{1}+r)}^{2} ={a_{1}(a_{1}+3r)}\\{ a_{1}}^{2}+2a_{1}r+r^{2}  ={ a_{1}}^{2}+3a_{1}r\\2a_{1}r+r^{2}  =3a_{1}r\\r(2a_{1}+r) =3a_{1}r\\r\neq0 \\deci\\2a_{1}+r =3a_{1}\\r =a_{1}\\a_{4} =a_{1}+3r=4r\\a_{6} =a_{1}+5r=6r\\a_{9} =a_{1}+8r=9r\\(a_{6})^{2} =a_{4}a_{9}\\cctd$