Question

у геометричній прогресії (bn) знайти: S10, якщо b1-b3=15, і b2-b4=30

Answer 1

$\displaystyle\bf b _{n} = b_{1}q {}^{n - 1} \\ \\ \left \{ {{b_{1} - b_{3} = 15} \atop {b_{2} - b_{4} = 30 }} \right. \\ \displaystyle\bf\\\left \{ {{b_{1} - b_{1} {q}^{2} = 15} \atop {b_{1}q - b_{1} {q}^{3} = 30}} \right. \\ \displaystyle\bf\\ \div \left \{ {{b_{1}(1 - {q}^{2} ) = 15} \atop {b_{1}q(1 - {q}^{2} ) = 30 }} \right. \\ \\ \frac{b_{1}(1 - {q}^{2}) }{b_{1}q( 1 - {q}^{2} )} = \frac{15}{30} \\ \frac{1}{q} = \frac{1}{2} \\ q = 2 \\ \\ b_{1} (1 - 2{}^{2} ) = 15 \\ b_{1}(1 - 4) = 15 \\ - 3b_{1} = 15 \\ b_{1} = 15 \div ( - 3) \\ b_{1} = - 5 \\ \\ S_{n} = \frac{b_{1}( {q}^{n} - 1) }{q - 1} \\ S_{10} = \frac{ - 5(2 {}^{10} - 1)}{2 - 1} = - 5(1024 - 1) = \\ = - 5 \times 1023 = - 5115$