Question

17. в убывающей арифметической прогрессии
a, +a,= 6
a, a, = 8
найдите S16
Ответ 20

Answer 1

Ответ:

q=

b

1

b

2

=

10

−8

=−0,8

�

=

�

1

1

−

�

=

10

1

−

(

−

0

,

8

)

=

10

1

,

8

=

50

9

=

5

5

9

S=

1−q

b

1

=

1−(−0,8)

10

=

1,8

10

=

9

50

=5

9

5

Answer 2

Ответ: S₁₆=-176.

Объяснение:

$\displaystyle\\\left \{ {{a_1+a_2=6} \atop {a_1*a_2=8}} \right. \ \ \ \ \ \ d < 0\ \ \ \ \ \ S_{16}=?\\\\\\\left \{ {{a_1+a_1+d=6} \atop {a_1*(a_1+d)=8}} \right. \ \ \ \ \ \ \ \left \{ {{2a_1+d=6} \atop {a_1^2+a_1d=8}} \right. \ \ \ \ \ \ \left \{ {{d=6-2a_1} \atop {a_1^2+a_1*(6-2a_1)=8}}\right. \\\\\\\left \{ {{d=6-2a_1} \atop {a_1^2+6a_1-2a_1^2=8}} \right. \ \ \ \ \ \ \left \{ {{d=6-2a_1} \atop {a_1^2-6a_1+8=0}} \right. \ \ \ \ \ \ \left \{ {{d=6-2a_1} \atop {a^2_1-4a_1-2a_1+8=0}} \right.$

$\displaystyle\\\left \{ {{d=6-2a_1} \atop {a_1*(a_1-4)-2*(a_1-4)=0}} \right. \ \ \ \ \ \ \left \{ {{d=6-2a_1} \atop {(a_1-4)*(a_1-2)=0}} \right. \\\\\\\left \{ {{d=6-2a_1} \atop {a_1=4\ \ \ \ a_2=2}} \right. \ \ \ \ \ \ \left \{ {{d_1=-2\ \ \ \ d_2=2\ \notin} \atop {a_1=4\ \ \ \ a_2=2}} \right. .\ \ \ \ \ \ \Rightarrow\\\\a_1=4\ \ \ \ d=-2.\\\\S_{16}=\frac{2*4+(16-1)*(-2)}{2}*16=(8-30)*8=-22*8=-176.$