Question

Найдите первый член арифметической прогрессии если а3=2√5-√3; а6=5√5-4√3

Answer 1

$left { {{a _{6}= a _{1} + 5d } atop {a _{3}= a _{1} + 2d }} right.$

$- left { {{5 sqrt{5}- 4 sqrt{3}= a _{1}+5d atop {2 sqrt{5}- sqrt{3} =a _{1}+2d }} right.$

$3d = 5 sqrt{5}-4 sqrt{3} - 2 sqrt{5}+ sqrt{3}$
$3d = 3 sqrt{5}-3 sqrt{3}$
$d = sqrt{5} - sqrt{3}$$a _{1} = a _{3} - 2d = 5 sqrt{5} - 4 sqrt{3} - 2 * ( sqrt{5}- sqrt{3}) = 5 sqrt{5 }- 4 sqrt{3} - 2 sqrt{5} +$$2 sqrt{3} = 3 sqrt{5} - 2 sqrt{3}$