Một công sai của cấp số cộng (un) thoả mãn: u1 + u15= 60; u4^2

* Hướng dẫn giải

Chọn  A

Ta  có: 

$$\begin{gathered} \left\{\begin{array}{l}{u}_7 + u_{15}= 60\\ u_4^2 +\hspace{0.17em}{u}_{11}^2= 981\end{array}\right.\iff \left\{\begin{array}{l}{u}_1+\hspace{0.17em}6d +\hspace{0.17em}{u}_1+\hspace{0.17em}14d = 60\\ {(u_1+\hspace{0.17em}3d)}^2+\hspace{0.17em}{(u_1+\hspace{0.17em}10d)}^2= 981\end{array}\right. \\ \iff \left\{\begin{array}{l}2u_1+\hspace{0.17em}20d = 60\\ u_1^2 +\hspace{0.17em}6u_{1}d +\hspace{0.17em}9d^2+\hspace{0.17em}{u}_1^2\hspace{0.17em}+ 20u_{1}d\hspace{0.17em}+\hspace{0.17em}100d^2= 981\end{array}\right. \\ \iff \left\{\begin{array}{l}{u}_1+\hspace{0.17em}10d = 30 \left(1\right)\hspace{0.17em}\\ 2u_1^2+\hspace{0.17em}26u_{1}d+\hspace{0.17em}109d^2= 981 \left(2\right)\hspace{0.17em}\end{array}\right. \end{gathered}$$

Từ (1);$u_1= 30- 10d$ thay vào(2) ta được:

$$\begin{gathered} 2. {(30-10 d)}^2+\hspace{0.17em}26. (30-10 d). d +\hspace{0.17em}109d^2= 981 \\ \iff 1800 - 1200d +\hspace{0.17em}200d^2+\hspace{0.17em}780d- 260d^2\hspace{0.17em}+\hspace{0.17em}109d^2- 981= 0 \\ \iff 49d^2- 420d +\hspace{0.17em}819 = 0 \\ \iff d = \frac{39}{7}; d = 3 \end{gathered}$$