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ЛизДжастис
@ЛизДжастис
August 2022
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Помогите решить, пожалуйста:
1. b3=27, b5=3, q=?
2. b1=3, q=1/3, S=?
3. bn= 9*(2/3)^n, S5=?
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mikael2
Verified answer
1/ b3=b1*q² b5=b1*q⁴ b5/b3=9=q² q=3 q=-3
2/ бесконечная геом. прогрессия s=b1/(1-q)=3:2/3=4.5
3/ bn+1/bn=9*(2/3)ⁿ⁺¹/9*(2/3)ⁿ=2/3=q b1=9*2/3=6
s5=6(1-(2/3)⁵):1/3=18*(1-32/125)=18*93/125
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Answers & Comments
Verified answer
1/ b3=b1*q² b5=b1*q⁴ b5/b3=9=q² q=3 q=-32/ бесконечная геом. прогрессия s=b1/(1-q)=3:2/3=4.5
3/ bn+1/bn=9*(2/3)ⁿ⁺¹/9*(2/3)ⁿ=2/3=q b1=9*2/3=6
s5=6(1-(2/3)⁵):1/3=18*(1-32/125)=18*93/125