ну да .......,...,.......... Created by vvvvttttrrrr666. matematika-ru

ну да .......,...,............

aalzhanov2010

Ответ:

$3)\ \ a_2=32\ ,\ \ a_6=20\ \ ,\ \ \ \ \ \boxed{\ a_{n}=a_1+d\, (n-1)\ }\\\\\\\left\{\begin{array}{l}a_1+d=32\\a_1+5d=20\end{array}\right\ \ominus \ \left\{\begin{array}{l}a_1+d=32\\-4d=12\end{array}\right\ \ \left\{\begin{array}{l}a_1=-d+32\\d=-3\end{array}\right\ \ \left\{\begin{array}{l}a_1=35\\d=-3\end{array}\right$

$a_n=35-3(n-1)=-13\ \ ,\ \ \ -3n+38=-13\ \ ,\ \ 3n=51\ \ ,\ \ \boldsymbol{n=17}\in N$

Получили натуральное число 17, которое является номером члена арифм. прогрессии, равного -13 .

$4)\ \ b_5=0,5\ \ ,\ \ b_7=0,005\qquad \qquad \boxed{\ b_{n}=b_1q^{n-1}\ }\\\\\\\left\{\begin{array}{l}b_1q^4=0,5\\b_1q^6=0,005\end{array}\right\ \ \left\{\begin{array}{l}b_1q^4=0,5\\\dfrac{b_1q^6}{b_1q^4}=\dfrac{0,005}{0,5}\end{array}\right\ \ \left\{\begin{array}{l}b_1q^4=0,5\\q^2=0,01\end{array}\right\ \ \left\{\begin{array}{l}b_1=\dfrac{0,5}{q^4}\\q=\pm 0,1\end{array}\right$

$a)\ \ \left\{\begin{array}{l}b_1=\dfrac{0,5}{q^4}\\q=-0,1\end{array}\right\ \ \left\{\begin{array}{l}b_1=\dfrac{0,5}{0,0001}\\q=-0,1\end{array}\right\ \ \left\{\begin{array}{l}b_1=5000\\q=-0,1\end{array}\right$

$b)\ \ a)\ \ \left\{\begin{array}{l}b_1=\dfrac{0,5}{q^4}\\q=0,1\end{array}\right\ \ \left\{\begin{array}{l}b_1=\dfrac{0,5}{0,0001}\\q=0,1\end{array}\right\ \ \left\{\begin{array}{l}b_1=5000\\q=0,1\end{array}\right\\\\\\Otvet:\ \ b_1=5000\ .$

$5)\ \ b_1+b_2+b_3=26\ \ ,\ \ \ q=3\ \ \ \qquad \boxed{\ b_{n}=b_1q^{n-1}\ }\\\\b_1+b_1q+b_1q^2=26\ \ ,\ \ \ \ b_1\cdot (1+q+q^2)=26\ \ ,\\\\\\b_1=\dfrac{26}{1+q+q^2}=\dfrac{26}{1+3+9}=\dfrac{26}{13}=2\\\\\\\boxed{S_{n}=\dfrac{b_1(1-q^{n})}{1-q}\ }\\\\\\S_6=\dfrac{2\cdot (1-3^6)}{1-3}=\dfrac{2\cdot (1-729)}{-2}=\boxed{\ 728\ }$

3 votes Thanks 3

NNNLLL54

Verified answer

Ответ:

1.Число 17 является

2.Первый член последовательности b₁=5000

3.Суммой первых шести членов прогрессии является 728

α₂ = 32 α₂ = α₁ + d

α₆ = 20 α₆ = α₁ + 5d

α₆ ₋ α₂ = α₁ + 5d ₋ α₁ - d = 4d

₋12 = 4d

d = 3

α₁ = α₂ ₋ d=32 - 3 = 35

35 + (n ₋ 1)d = -13

35 - 3n + 3 = ₋13

3n = 51, n = 17

b₇ = b₅ · q · q = b₅· q₂

0 . 005 = 0 . 5 · q₂

q² = 0.005 ÷ 0.5

q² = 0.01

q₁ = 0.1

q₂ = -0.1

1) q=0.1

b₅ = b₁ · q₄

b₁ = b₅/q₄ = 0.5/0.1₄ = 5000

2) q = ₋0.1

b₁ = b₅/q₄ = 0.5/(₋0.1)₄ = 5000

b1 + b1q + b1q₂ = 26

b1 + 3b1 + 9b1 = 26

13b1 = 26

b1 = 2

S6 = b1 · (1-q₆)/(1-q)

S = 2 · (1-729)/(₋2) = 728

3 votes Thanks 1

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