Question

в геометрической прогрессии b1+b2=21 , а b2+b3=42, найдите b1+b5

Answer 1

Ответ:

$\left\{\begin{array}{l}b_1+b_2=21\\b_2+b_3=42\end{array}\right\ \ \left\{\begin{array}{l}b_1+b_1q=21\\b_1q+b_1q^2=42\end{array}\right\ \ \left\{\begin{array}{l}b_1(1+q)=21\\b_1q(1+q)=42\end{array}\right\ \ \left\{\begin{array}{l}b_1(1+q)=21\\21\cdot q=42\end{array}\right\\\\\\\left\{\begin{array}{l}b_1\cdot (1+2)=21\\q=2\end{array}\right\ \ \left\{\begin{array}{l}b_1=7\\q=2\end{array}\right$

$b_1+b_5=b_1+b_1q^4=b_1\cdot (1+q^4)=7\cdot (1+2^4)=7\cdot 17=119$