1. ## Geometric progression?

If $\displaystyle a_{i}\in\mathbb{R}, i = 1,2,3,...N$and all $\displaystyle a_{i}\ 's$ are distinct such that:
$\displaystyle \left(\sum_{i = 1}^{n - 1} a_{i}^2\right)x^2 + 2\left(\sum_{i = 1}^{n - 1} a_{i}a_{i + 1}\right)x + \sum_{i = 2}^{n} a_{i}^2\leq 0$
,then show that $\displaystyle a_{1},a_{2},a_{3}...$ are in G.P.

2. Originally Posted by fardeen_gen
If $\displaystyle a_{i}\in\mathbb{R}, i = 1,2,3,...N$and all $\displaystyle a_{i}\ 's$ are distinct such that:
$\displaystyle \left(\sum_{i = 1}^{n - 1} a_{i}^2\right)x^2 + 2\left(\sum_{i = 1}^{n - 1} a_{i}a_{i + 1}\right)x + \sum_{i = 2}^{n} a_{i}^2\leq 0$
,then show that $\displaystyle a_{1},a_{2},a_{3}...$ are in G.P.
Hi fardeen_gen.

The discriminant is

$\displaystyle 4\left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)^2-4\left(\sum_{i\,=\,1}^{n-1}a_i^2\right)\left(\sum_{i\,=\,2}^na_i^2\right)$

which is never positive by the Cauchy–Schwarz inequality. Hence, since the coefficient of $\displaystyle x^2$ is never negative, we have

$\displaystyle \left(\sum_{i\,=\,1}^{n-1}a_i^2\right)x^2+2\left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)x+\left(\sum_{i\,=\,2}^{n-1}a_i^2\right)\ \ge\ 0$

Combining this with the given condition, we find that the quadratic expression is equal to zero. Hence the discriminant is zero, so we have equality for Cauchy–Schwarz:

$\displaystyle \left(\sum_{i\,=\,1}^{n-1}a_ia_{i+1}\right)^2\ =\ \left(\sum_{i\,=\,1}^{n-1}a_i^2\right)\left(\sum_{i\,=\,2}^na_i^2\right)$

This means that the vectors $\displaystyle \left(a_1,a_2,\ldots,a_{n-1}\right)$ and $\displaystyle \left(a_2,a_3,\ldots,a_n\right)$ are linearly dependent, i.e. $\displaystyle \exists\,r$ such that $\displaystyle r\left(a_1,a_2,\ldots,a_{n-1}\right)=\left(a_2,a_3,\ldots,a_n\right).$ Hence the $\displaystyle a_i$’s are in geometric progression.