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Petr8433
@Petr8433
July 2022
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Найдите сумму десяти первых членов геометрической прогрессии (xn) если X1= 0,48 , x2=0,32
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IrkaShevko
Verified answer
Q = x2/x1 = 0,32/0,48 = 2/3
Sn = b1(qⁿ - 1)/(q-1)
S10 = 0,48((2/3)¹⁰ - 1)/(2/3 - 1) = 1,44(1 - 1024/59049) = 1,44*58025/59049= 83556/59049 = 1 24507/59049 = 1 2723/6561
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Answers & Comments
Verified answer
Q = x2/x1 = 0,32/0,48 = 2/3Sn = b1(qⁿ - 1)/(q-1)
S10 = 0,48((2/3)¹⁰ - 1)/(2/3 - 1) = 1,44(1 - 1024/59049) = 1,44*58025/59049= 83556/59049 = 1 24507/59049 = 1 2723/6561