Question

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найдите сумму первых шести членов геометрической прогрессии если b2=9,q=-1/3

Answer 1

$\\b_n = b_1 * q^{n-1}\\ \\b_2 = b_1*q\\ \\b_1 = \frac{b_2}{q}\\ \\b_1 = \frac{9}{-\frac{1}{3}} = -27\\ \\b_6 = b_1 * q^5\\ \\b_6 = -27 * (-\frac{1}{3})^5 = -3^3 * (-\frac{1^5}{3^5} = \frac{1}{3^2} = \frac{1}{9}\\ \\S_n = \frac{b_n*q-b_1}{q-1} ; q \neq 1\\ \\S_6 = \frac{b_6*q-b_1}{q-1};q \neq 1$
$\\S_6 = \frac{\frac{1}{9} * (-\frac{1}{3} - (-27)}{-\frac{1}{3} - 1}\\ = \\ = \frac{-\frac{1}{27} + 27}{-\frac{4}{3} }= (-\frac{1}{27} + 27)* - \frac{3}{4} =\\ = -\frac{1}{27} * (-\frac{3}{4}) + 27*(-\frac{3}{4}) = \\ = \frac{1}{36} - \frac{81}{4} = -(\frac{729}{36} - \frac{1}{36}) = - 20 \frac{8}{36} = - 20\frac{2}{9}$