"If $a_1,a_2,a_3,\dots,a_n$ are distinct non-zero numbers such that
$$ \left(\sum_{k=1}^{n-1} {{a_k}^2}\right)x^2 +2\left(\sum_{k=1}^{n-1} a_k a_{k+1}\right)x + \left(\sum_{k=2}^{n} {{a_k}^2}\right)\le 0 $$"
We have to tell the kind of sequence that $a_1,a_2,a_3,\dots,a_n$ produces. (Options are AP, GP, HP or AGP).
Since the coefficient of $x^2$ is positive and it says that the function is less than or equal to zero, that means the parabola is opening upwards (towards positive Y-axis) and it at least touches the X-axis. Thus I tried this: the discriminant $D \ge 0$ but things are getting quite messy.
Can anyone help me figure it out?
From Cauchy-Schwarz inequality, $(2\sum\limits_{k=1}^{n-1}a_k a_{k+1})^2-4\sum\limits_{k=1}^{n-1}a_{k}^2\sum\limits_{k=2}^n a_{k}^2\leq 0$, so: $(\sum\limits_{k=1}^{n-1}a_{k}^2) x^2+(2\sum\limits_{k=1}^{n-1}a_k a_{k+1})x+ \sum\limits_{k=2}^n a_{k}^2 =0$ has only one solution. We can get $(2\sum\limits_{k=1}^{n-1}a_k a_{k+1})^2-4\sum\limits_{k=1}^{n-1}a_{k}^2\sum\limits_{k=2}^n a_{k}^2=0$, then $\frac{a_{k+1}}{a_k}=c$, here $c$ is a constant.