Question

Найдите сумму первых шести членов геометрической прогрессии 54;36

Answer 1

Ответ:

b1=54 ; b2=36

$q=\frac{b1}{b2} =\frac{54}{36} =\frac{2}{3}$

$S_{n}=\frac{b1(q^n-1)}{q-1}$   =>

$S_{6}=\frac{54((\frac{2}{3})^6 -1)}{\frac{2}{3}-1} = \frac{54(\frac{64}{729}-1 )}{-\frac{1}{3} } = \frac{54 * (-\frac{665}{729} )}{-\frac{1}{3}}=\frac{-\frac{35910}{729} }{-\frac{1}{3}}=\\ =\frac{35910}{729} : \frac{1}{3} = \frac{1330}{27} * \frac{3}{1}= \frac{3990}{27}$

$S_{6} = \frac{3990}{27}= \frac{1330}{9} = 147\frac{7}{9}$