Найдите первый член геометрической прогрессии,в которой q=2/3,S4=65

65 =b₁(2/3⁴-1) /(2/3-1) ;65*(-1/3) = b₁(16/81 -1) ;65*(-1/3) = b₁*( -65/81) * * * ⇔1 =b₁/27 .* * *b₁ = 65*(-1/3) : ( -65/81) =65*(-1/3) *(-81)/ 65 =(1/3)*81 =27.

3 +3*(2/3) +3*(2/3)² +3*(2/3)³ +3*(2/3)⁴ =3 +2 +4/3 +8/9 +16/27 =211/27 =7 22/27 ≠ 65.

 Инфляция баллов.-------------------------S₄ =b₁(1-q⁴)/(1-q) ⇒b₁ =S₄(1-q)/(1-q⁴) ;b₁ =65(1-2/3)/(1-2/3)⁴) =(65/3)/(65/81)81/3=27.