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alisher270502
@alisher270502
August 2022
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b(1)+b(2)+b(3)=35
b(1)^2+b(2)^2+b(3)^2=525 геометрическая прогресия
b(1)-? q-?
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IrkaShevko
Verified answer
B2 = b1*q
b3 = b1 * q²
b1(1 + q + q²) = 35
b1²(1 + q² + q⁴) = 525
возведем первое в квадрат
b1²(1 + q² + q⁴ + 2q + 2q² + 2q³) = 1225
b1²(1 + q² + q⁴) = 525
b1²(1 + q² + q⁴) + 2q*b1²(1 + q + q²) = 1225
525 + 2q*b1*35 = 1225
70q*b1 = 700
q*b1 = 10
b1 = 10/q
10(1 + q + q²)/q = 35
2 + 2q + 2q² = 7q
2q² - 5q + 2 = 0
D = 25 - 16 = 9
q = 1/2 или q = 2
b1 = 20 или b1 = 5
Ответ: b1 = 20, q = 0,5 или b1 = 5, q = 2
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Answers & Comments
Verified answer
B2 = b1*qb3 = b1 * q²
b1(1 + q + q²) = 35
b1²(1 + q² + q⁴) = 525
возведем первое в квадрат
b1²(1 + q² + q⁴ + 2q + 2q² + 2q³) = 1225
b1²(1 + q² + q⁴) = 525
b1²(1 + q² + q⁴) + 2q*b1²(1 + q + q²) = 1225
525 + 2q*b1*35 = 1225
70q*b1 = 700
q*b1 = 10
b1 = 10/q
10(1 + q + q²)/q = 35
2 + 2q + 2q² = 7q
2q² - 5q + 2 = 0
D = 25 - 16 = 9
q = 1/2 или q = 2
b1 = 20 или b1 = 5
Ответ: b1 = 20, q = 0,5 или b1 = 5, q = 2