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алина3893
алина3893
02.06.2022 06:46 •  Алгебра

Дано геометричну прогресію (bn), у якій b1=10,q=1/2, bn=1,25. Знайдіть номер n. А) n=3
Б) n=2
В) n=5
Г) n=4

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Ответ:
kapitoshka202
kapitoshka202
02.06.2022

Б

4,4(22 оценок)
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Ответ:
ksushakorovina1
ksushakorovina1
02.06.2022

Объяснение:

1.

a8=a7+d

d=a8

a8=a7+a8=>a7=0

2.

a1 = -12

a2 = -9

an = a1 + d * (n - 1);

a2 = a1 + d;

a2 - a1 = d.

d = -9 - (-12) = 3.

a8 = a1 + 7 * d;

a8 = -12 + 7 * 3;

a8 = 9.

S8 = (a1 + a8) * 8/2;

S8 = 4 * (-12 + 9);

S8 = -12.

3.

A6=a1+d(6-1), a7=a1+d(7-1), a11=a1+d(11-1), a12=a1+d(12-1).

(a1+6d)+(a1+d11)+8=(a1+5d)+(a1+10d)

a1+6d+a1+11d+8=aq+5d+a1+10

17d+8=15d

2d=-8

d=-4

4.

q=4/12=1/3

b9=12/1/3=36

5.

a1=a3:q²

a1=36:9

a1=4

s5=a1.q^4

s5=4.3^4, s5=4.81, s5=324

6.

a8=a7*q=a7*a8

a7=a8/a8=1

7.

A5=a1*q^4

Q^4=5

A13=a1*q^12=a5*q^8

A13/a5=q8=25

8.

an = a1 + (n - 1)d;

an = 6 + 4(n - 1);

an > 260;

6 + 4(n - 1) > 260;

4(n - 1) > 260 - 6;

4(n - 1) > 254;

n - 1 > 254/4;

n - 1 > 63,5;

n > 63,5 + 1;

n > 64,5;

9.

A1=6

a6=17

a2, a3, a4, a5-?

a6=a1+5d

d = (a5-a1) / 5

d = (17-6) / 5=11/5=2,2

a2=a1+d=6+2,2=8,2

a3=a2+d=8,2+2,2=10,4

a4=a3+d=10,4+2,2=12,6

a5=a4+d=12,6+2,2=14,8

10.

а1=60

аn=110

N=51

(2*60+50)*51/2=4335

11.

Sn = b1 * (1 - qn)/(1 - q).

S4 = b1 * (1 - (- 3)4)/(1 - (- 3)) = - 40.

b1 = (- 40) : (1 - 81)/(1 + 3) = - 40 * 4/(- 80) = 2.

S8 = b1 * (1 - (- 3)8)/(1 - (- 3)) = 2 * (1 - 6561)/4 = - 6560/2 = - 3280.

13.

аn=1+7*(n-1)=1+7n-7= 7n-6

28+6=34

55+6=61

9156:7=1308

14.

a2=a1+d; 4=a1+d

a28=a1+27d; 56=a1+27d

a28-a2=56-4=52

52=26d

d=2

S28=(2a1+d(n-1))/2 s=(4+54)/2=29

a2=4=a1+d,то a1=2

15.

a6=a1+5d

a10=a1+9d

a16=a1+15d

а10-а6=4d

а10-а6=20-14=6

d=1.5

а16 и а10:

а16-а10=6d

28-20=8

d=8/6=4/3

d разные получаются - значит числа не принадлежат арифметической прогрессии

4,4(5 оценок)
Ответ:
fhctybq123
fhctybq123
02.06.2022
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4,4(31 оценок)
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