Question

Задание 12. Найти сумму пяти первых членов геометрической прогрессии, в которой b2 + b3 = 18, b4 - bz = 18. [4]​

Answer 1

$\displaystyle\bf b_{n} = b_{1} {q}^{n - 1} \\ \\\left \{ {{b_{2} + b_{3} = 18} \atop {b_{4} - b_{2} = 1 8}} \right. \\ \displaystyle\bf\\\left \{ {{b_{1}q + b_{1} {q}^{2} = 18} \atop {b_{1} {q}^{3} - b_{1}q = 18}} \right. \\ \displaystyle\bf\\ \div \left \{ {{b_{1}q(1 + q) = 18} \atop {b_{1}q( {q}^{2} - 1) = 18}} \right. \\ \\ \frac{1 + q}{q {}^{2} - 1} = 1 \\ \frac{1 + q}{(q - 1)(q + 1)} = 1 \\ \frac{1}{q - 1} = 1 \\ q - 1 = 1 \\ q = 1 + 1 \\ q = 2 \\ \\ b_{1} \times 2 + b _{1} \times 2 {}^{2} =18 \\ 2b _{1} + 4b_{1} = 18 \\ 6b_{1} = 18 \\ b_{1} = 18 \div 6 \\ b_{1} = 3 \\ \\ b_{5} = b_{1} {q}^{4} = 3 \times {2}^{4} = 3 \times 16 = 48 \\ S_{n} = \frac{b_{n}q + b_{1}}{q - 1} \\ S_{5} = \frac{b_{5}q + b_{1}}{q - 1} = \frac{48 \times 2 + 3}{2 - 1} = 96 + 3 = 101$