Question

Cho cấp số nhân thoả mãn: \(\left\{{}\begin{matrix}u_5-u_1=15\\u_4-u_2=6\end{matrix}\right.\). Tính \(u_1\) ?

Answer 1

\(\left\{{}\begin{matrix}u_5-u_1=15\\u_4-u_2=6\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}u_1.q^4-u_1=15\\u_1.q^3-u_1.q=6\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}u_1\left(q^4-1\right)=15\\u_1\left(q^3-q\right)=6\end{matrix}\right.\)

⇔ \(\dfrac{q^4-1}{q^3-q}\) = \(\dfrac{15}{6}\) ⇔ \(\dfrac{\left(q^2-1\right)\left(q^2+1\right)}{q\left(q^2-1\right)}\) = \(\dfrac{5}{2}\)

⇔ \(2q^2\)+2 - \(5q\)= 0

⇔\(\left[{}\begin{matrix}q=2\rightarrow u_1=1\\q=\dfrac{1}{2}\rightarrow u_1=-16\end{matrix}\right.\)